1.7: Properties of Derivatives

We will now develop some properties of derivatives with the aim of facilitating their calculation for certain general classes of functions.

To begin, if \(f(x)=k\) for all \(x\) and some real constant \(k,\) then, for any infinitesimal \(d x,\) \[f(x+d x)-f(x)=k-k=0 .\] Hence, if \(d x \neq 0\), \[\frac{f(x+d x)-f(x)}{d x}=0 ,\] and so \(f^{\prime}(x)=0 .\) In other words, the derivative of a constant is \(0 .\)

Theorem \(\PageIndex{1}\)

For any real constant \(k\),

\[\frac{d}{d x} k=0 .\]

Example \(\PageIndex{1}\)

\[\frac{d}{d x} 4=0.\]

Sums and Differences

Now suppose \(u\) and \(v\) are both differentiable functions of \(x .\) Then, for any infinitesimal \(d x\), \[\begin{aligned} d(u+v) &=(u(x+d x)+v(x+d x))-(u(x)-v(x)) \\ &=(u(x+d x)-u(x))+(v(x+d x)-v(x)) \\ &=d u+d v. \end{aligned}\] Hence, if \(d x \neq 0\), \[\frac{d(u+v)}{d x}=\frac{d u}{d x}+\frac{d v}{d x}.\] In other words, the derivative of a sum is the sum of the derivatives.

Theorem \(\PageIndex{2}\)

If \(f\) and \(g\) are both differentiable and \(s(x)=f(x)+g(x)\), then

\[s^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x) .\]

Example \(\PageIndex{2}\)

If \(y=x^{2}+\sqrt{x},\) then, using our results from the previous section,

\[\frac{d y}{d x}=\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(\sqrt{x})=2 x+\frac{1}{2 \sqrt{x}} .\] A similar argument shows that \[\frac{d}{d x}(u-v)=\frac{d u}{d x}-\frac{d v}{d x} .\]

Exercise \(\PageIndex{1}\)

Find the derivative of \(y=x^{2}+5\).

Answer

\(\frac{d y}{d x}=2 x\)

Exercise \(\PageIndex{2}\)

Find the derivative of \(f(x)=\sqrt{x}-x^{2}+3\).

Answer

\(f^{\prime}(x)=\frac{1}{2 \sqrt{x}}-2 x\)

Constant Multiples

If \(c\) is any real constant and \(u\) is a differentiable function of \(x,\) then, for any infinitesimal \(d x,\)

\[d(c u)=c u(x+d x)-c u(x)=c(u(x+d x)-u(x))=c d u .\] Hence, if \(d x \neq 0\), \[\frac{d(c u)}{d x}=c \frac{d u}{d x},\] In other words, the derivative of a constant times a function is the constant times the derivative of the function.

Theorem \(\PageIndex{3}\)

If \(c\) is a real constant, \(f\) is differentiable, and \(g(x)=c f(x)\), then

\[g^{\prime}(x)=c f^{\prime}(x).\]

Example \(\PageIndex{3}\)

If \(y=5 x^{2},\) then

\[\frac{d y}{d x}=5 \frac{d}{d x}\left(x^{2}\right)=5(2 x)=10 x .\]

Exercise \(\PageIndex{3}\)

Find the derivative of \(y=8 x^{2}\).

Answer

\(\frac{d y}{d x}=16 x\)

Exercise \(\PageIndex{4}\)

Find the derivative of \(f(x)=4 \sqrt{x}+15\).

Answer

\(f^{\prime}(x)=\frac{2}{\sqrt{x}}\)

Products

Again suppose \(u\) and \(v\) are differentiable functions of \(x\). Note that, in particular, \(u\) and \(v\) are continuous, and so both \(d u\) and \(d v\) are infinitesimal for any infinitesimal \(d x .\) Moreover, note that

\[u(x+d x)=u(x)+d u \text { and } v(x+d x)=v(x)+d v .\] Hence \[\begin{aligned} d(u v) &=u(x+d x) v(x+d x)-u(x) v(x) \\ &=(u(x)+d u)(v(x)+d v)-u(x) v(x) \\ &=(u(x) v(x)+u(x) d v+v(x) d u+d u d v)-u(x) v(x) \\ &=u d v+v d u+d u d v, \end{aligned}\] and so, if \(d x \neq 0\), \[\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}+d u \frac{d v}{d x} \simeq u \frac{d v}{d x}+v \frac{d u}{d x}\] Thus we have, for any differentiable functions \(u\) and \(v\), \[\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x} ,\] which we call the product rule.

Theorem \(\PageIndex{4}\)

If \(f\) and \(g\) are both differentiable and \(p(x)=f(x) g(x),\) then

\[p^{\prime}(x)=f(x) g^{\prime}(x)+g(x) f^{\prime}(x) .\]

Example \(\PageIndex{4}\)

We may use the product rule to find a formula for the derivative of a positive integer power of \(x .\) We first note that if \(y=x,\) then, for any infinitesimal \(d x,\)

\[d y=(x+d x)-x=d x ,\] and so, if \(d x \neq 0\), \[\frac{d y}{d x}=\frac{d x}{d x}=1 .\] Thus we have \[\frac{d}{d x} x=1 ,\] as we should expect, since \(y=x\) implies that \(y\) changes at exactly the same rate as \(x .\) Using the product rule, it now follows that \[\frac{d}{d x} x^{2}=\frac{d}{d x}(x \cdot x)=x \frac{d}{d x} x+x \frac{d}{d x} x=x+x=2 x ,\] in agreement with a previous example. Next, we have \[\frac{d}{d x} x^{3}=x \frac{d}{d x} x^{2}+x^{2} \frac{d}{d x} x=2 x^{2}+x^{2}=3 x^{2}\] and \[\frac{d}{d x} x^{4}=x \frac{d}{d x} x^{3}+x^{3} \frac{d}{d x} x=3 x^{3}+x^{3}=4 x^{3} .\] At this point we might suspect that for any integer \(n \geq 1\), \[\frac{d}{d x} x^{n}=n x^{n-1} .\] This is in fact true, and follows easily from an inductive argument: Suppose we have shown that for any \(k<n\), \[\frac{d}{d x} x^{k}=k x^{k-1} .\] Then \[\begin{aligned} \frac{d}{d x} x^{n} &=x \frac{d}{d x} x^{n-1}+x^{n-1} \frac{d}{d x} x \\ &=x\left((n-1) x^{n-2}\right)+x^{n-1} \\ &=n x^{n-1} . \end{aligned}\] We call this result the power rule.

Theorem \(\PageIndex{5}\)

For any integer \(n \geq 1\),

\[\frac{d}{d x} x^{n}=n x^{n-1} .\] We shall see eventually, in Theorems \(1.7 .7,1.7 .10,\) and \(2.7 .2,\) that the power rule in fact holds for any real number \(n \neq 0 .\)

Example \(\PageIndex{5}\)

When \(n=34,\) the power rule shows that

\[\frac{d}{d x} x^{34}=34 x^{33} .\]

Example \(\PageIndex{6}\)

If \(f(x)=14 x^{5},\) then, combining the power rule with our result for constant multiples,

\[f^{\prime}(x)=14\left(5 x^{4}\right)=70 x^{4} .\]

Exercise \(\PageIndex{5}\)

Find the derivative of \(y=13 x^{5}\).

Answer

\(\frac{d y}{d x}=65 x^{4}\)

Example \(\PageIndex{7}\)

Combining the power rule with our results for constant multiples and differences, we have

\[\frac{d}{d x}\left(3 x^{2}-5 x\right)=6 x-5 .\]

Exercise \(\PageIndex{6}\)

Find the derivative of \(f(x)=5 x^{4}-3 x^{2}\).

Answer

\(f^{\prime}(x)=20 x^{3}-6 x\)

Exercise \(\PageIndex{7}\)

Find the derivative of \(y=3 x^{7}-3 x+1\).

Answer

\(\frac{d y}{d x}=21 x^{6}-3\)

Polynomials

As the previous examples illustrate, we may put together the above results to easily differentiate any polynomial function. That is, if \(n \geq 1\) and \(a_{n}, a_{n-1}\), \(\ldots, a_{0}\) are any real constants, then

\[\begin{aligned} \frac{d}{d x}\left(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{2} x^{2}+a_{1} x+a_{0}\right) \\=n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\cdots+2 a_{2} x+a_{1} .\end{aligned}\]

Example \(\PageIndex{8}\)

If \(p(x)=4 x^{7}-13 x^{3}-x^{2}+21,\) then

\[p^{\prime}(x)=28 x^{6}-39 x^{2}-2 x .\]

Exercise \(\PageIndex{8}\)

Find the derivative of \(f(x)=3 x^{5}-6 x^{4}-5 x^{2}+13\).

Answer

\(f^{\prime}(x)=15 x^{4}-24 x^{3}-10 x\)

Quotients

If \(u\) is a differentiable function of \(x, u(x) \neq 0,\) and \(d x\) is an infinitesimal, then

\[\begin{aligned} d\left(\frac{1}{u}\right) &=\frac{1}{u(x+d x)}-\frac{1}{u(x)} \\ &=\frac{1}{u(x)+d u}-\frac{1}{u(x)} \\ &=\frac{u-(u+d u)}{u(u+d u)} \\ &=\frac{-d u}{u(u+d u)} . \end{aligned}\] Hence, since \(u+d u \simeq u,\) if \(d x \neq 0\), \[\frac{d}{d x}\left(\frac{1}{u}\right)=-\frac{\frac{d u}{d x}}{u(u+d u)} \simeq-\frac{1}{u^{2}} \frac{d u}{d x} .\]

Theorem \(\PageIndex{6}\)

If \(f\) is differentiable, \(f(x) \neq 0,\) and

\[g(x)=\frac{1}{f(x)} ,\] then \[g^{\prime}(x)=-\frac{f^{\prime}(x)}{f(x)^{2}} .\]

Example \(\PageIndex{9}\)

If

\[f(x)=\frac{1}{x^{2}} ,\] then \[f^{\prime}(x)=-\frac{1}{x^{4}} \cdot 2 x=-\frac{2}{x^{3}} .\] Note that the result of the previous example is the same as we would have obtained from applying the power rule with \(n=-2\). In fact, we may now show that the power rule holds in general for negative integer powers: If \(n<0\) is an integer, then \[\frac{d}{d x} x^{n}=\frac{d}{d x}\left(\frac{1}{x^{-n}}\right)=-\frac{1}{x^{-2 n}} \cdot\left(-n x^{-n-1}\right)=n x^{n-1} .\] Hence we now have our first generalization of the power rule.

Theorem \(\PageIndex{7}\)

For any integer \(n \neq 0\),

\[\frac{d}{d x} x^{n}=n x^{n-1} .\]

Example \(\PageIndex{10}\)

If

\[f(x)=3 x^{2}-\frac{5}{x^{7}} ,\] then \(f(x)=3 x^{2}-5 x^{-7},\) and so \[f^{\prime}(x)=6 x+35 x^{-8}=6 x+\frac{35}{x^{8}} .\] Now suppose \(u\) and \(v\) are both differentiable functions of \(x\) and let \[y=\frac{u}{v} .\] Then \(u=v y,\) so, as we saw above, \[d u=v d y+y d v+d v d y=y d v+(v+d v) d y .\] Hence, provided \(v(x) \neq 0\), \[d y=\frac{d u-y d v}{v+d v}=\frac{d u-\frac{u}{v} d v}{v+d v}=\frac{v d u-u d v}{v(v+d v)} .\] Thus, for any nonzero infinitesimal \(d x\), \[\frac{d y}{d x}=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v(v+d v)} \simeq \frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} .\] This is the quotient rule.

Theorem \(\PageIndex{8}\)

If \(f\) and \(g\) are differentiable, \(g(x) \neq 0,\) and

\[q(x)=\frac{f(x)}{g(x)} ,\] then \[q^{\prime}(x)=\frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{(g(x))^{2}} .\] One consequence of the quotient rule is that, since we already know how to differentiate polynomials, we may now differentiate any rational function easily.

Example \(\PageIndex{11}\)

If

\[f(x)=\frac{3 x^{2}-6 x+4}{x^{2}+1} ,\] then \[\begin{aligned} f^{\prime}(x) &=\frac{\left(x^{2}+1\right)(6 x-6)-\left(3 x^{2}-6 x+4\right)(2 x)}{\left(x^{2}+1\right)^{2}} \\ &=\frac{6 x^{3}-6 x^{2}+6 x-6-6 x^{3}+12 x^{2}-8 x}{\left(x^{2}+1\right)^{2}} \\ &=\frac{6 x^{2}-2 x-6}{\left(x^{2}+1\right)^{2}} . \end{aligned}\]

Example \(\PageIndex{12}\)

We may use either 1.7.30 or 1.7.37 to differentiate

\[y=\frac{5}{x^{2}+1} .\] In either case, we obtain \[\frac{d y}{d x}=-\frac{5}{\left(x^{2}+1\right)^{2}} \frac{d}{d x}\left(x^{2}+1\right)=-\frac{10 x}{\left(x^{2}+1\right)^{2}} .\]

Exercise \(\PageIndex{9}\)

Find the derivative of

\[y=\frac{14}{4 x^{3}-3 x} .\]

Answer

\(\frac{d y}{d x}=\frac{42-168 x^{2}}{\left(4 x^{3}-3 x\right)^{2}}\)

Exercise \(\PageIndex{10}\)

Find the derivative of

\[f(x)=\frac{4 x^{3}-1}{x^{2}-5} .\]

Answer

\(f^{\prime}(x) \frac{4 x^{4}-60 x^{2}+2 x}{\left(x^{2}-5\right)^{2}}\)

Composition of Functions

Suppose \(y\) is a differentiable function of \(u\) and \(u\) is a differentiable function of \(x .\) Then \(y\) is both a function of \(u\) and a function of \(x,\) and so we may ask for the derivative of \(y\) with respect to \(x\) as well as the derivative of \(y\) with respect to \(u .\) Now if \(d x\) is an infinitesimal, then

\[d u=u(x+d x)-u(x)\] is also an infinitesimal (since \(u\) is continuous). If \(d u \neq 0\), then the derivative of \(y\) with respect to \(u\) is equal to the shadow of \(\frac{d y}{d u}\). At the same time, if \(d x \neq 0\), the derivative of \(u\) with respect to \(x\) is equal to the shadow of \(\frac{d u}{d x} .\) But \[\frac{d y}{d u} \frac{d u}{d x}=\frac{d y}{d x} ,\] and the shadow of \(\frac{d y}{d x}\) is the derivative of \(y\) with respect to \(x .\) It follows that the derivative of \(y\) with respect to \(x\) is the product of the derivative of \(y\) with respect to \(u\) and the derivative of \(u\) with respect to \(x\). Of course, \(d u\) is not necessarily nonzero even if \(d x \neq 0\) (for example, if \(u\) is a constant function), but the result holds nevertheless, although we will not go into the technical details here. We call this result the chain rule.

Theorem \(\PageIndex{9}\)

If \(y\) is a differentiable function of \(u\) and \(u\) is a differentiable function of \(x,\) then

\[\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x} .\] Not that if we let \(y=f(u), u=g(x),\) and \[h(x)=f \circ g(x)=f(g(x)) ,\] then \[\frac{d y}{d x}=h^{\prime}(x), \frac{d y}{d u}=f^{\prime}(g(x)), \text { and } \frac{d u}{d x}=g^{\prime}(x) .\] Hence we may also express the chain rule in the form \[h^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x) .\]

Example \(\PageIndex{13}\)

If \(y=3 u^{2}\) and \(u=2 x+1,\) then

\[\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}=(6 u)(2)=12 u=24 x+12 .\] We may verify this result by first finding \(y\) directly in terms of \(x,\) namely, \[y=3 u^{2}=3(2 x+1)^{2}=3\left(4 x^{2}+4 x+1\right)=12 x^{2}+12 x+3 ,\] and then differentiating directly: \[\frac{d y}{d x}=\frac{d}{d x}\left(12 x^{2}+12 x+3\right)=24 x+12 .\] Note that if we want to evaluate \(\frac{d y}{d x}\) when, for example, \(x=2,\) we may either evaluate the final form, that is, \[\left.\frac{d y}{d x}\right|_{x=2}=\left.(24 x+12)\right|_{x=2}=48+12=60 ,\] or, noting that \(u=5\) when \(x=2,\) the intermediate form, that is, \[\left.\frac{d y}{d x}\right|_{x=2}=\left.12 u\right|_{u=5}=60 .\] In other words, \[\left.\frac{d y}{d x}\right|_{x=2}=\left.\left.\frac{d y}{d u}\right|_{u=5} \frac{d u}{d x}\right|_{x=2} .\]

Exercise \(\PageIndex{11}\)

If \(y=u^{3}+5\) and \(u=x^{2}-1,\) find \(\left.\frac{d y}{d x}\right|_{x=1}\).

Answer

\(\left.\frac{d y}{d x}\right|_{x=1}=216\)

Example \(\PageIndex{14}\)

If \(h(x)=\sqrt{x^{2}+1},\) then \(h(x)=f(g(x))\) where \(f(x)=\sqrt{x}\) and \(g(x)=x^{2}+1 .\) Since

\[f^{\prime}(x)=\frac{1}{2 \sqrt{x}} \text { and } g^{\prime}(x)=2 x ,\] it follows that \[h^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x=\frac{x}{\sqrt{x^{2}+1}} .\]

Exercise \(\PageIndex{12}\)

Find the derivative of \(f(x)=\sqrt{4 x+6}\).

Answer

\(f^{\prime}(x)=\frac{2}{\sqrt{4 x+6}}\)

Exercise \(\PageIndex{13}\)

Find the derivative of \(y=\left(x^{2}+5\right)^{10}\).

Answer

\(\frac{d y}{d x}=20 x\left(x^{2}+5\right)^{9}\)

Example \(\PageIndex{15}\)

As we saw in Example \(1.2 .5,\) if \(M\) is the mass, in grams, of a spherical balloon being filled with water and \(r\) is the radius of the balloon, in centimeters, then

\[M=\frac{4}{3} \pi r^{3} \text { grams } \] and \[\frac{d M}{d r}=4 \pi r^{2} \text { grams / centimeter } ,\] a result which we may verify easily now using the power rule. Suppose water is being pumped into the balloon so that the radius of the balloon is increasing at the rate of 0.1 centimeters per second when the balloon has a radius of 10 centimeters. since \(M\) is a function of time \(t,\) as well as a function of the radius of the balloon \(r,\) we might wish to know the rate of change of \(M\) with respect to \(t .\) since we are given that \[\left.\frac{d r}{d t}\right|_{r=10}=0.1 \text { centimeters/second } ,\] we may use the chain rule to find that \[\left.\frac{d M}{d t}\right|_{r=10}=\left.\left.\frac{d M}{d r}\right|_{r=10} \frac{d r}{d t}\right|_{r=10}=(400 \pi)(0.1)=40 \pi \text { grams } / \text { second } .\]

Exercise \(\PageIndex{14}\)

Suppose \(A\) is the area and \(r\) is the radius of a circular wave at time \(t .\) Suppose when \(r=100\) centimeters the radius of the circle is increasing at a rate of 2 centimeters per second. Find the rate at which the area of the circle is growing when \(r=100\) centimeters.

Answer

\(\left.\frac{d A}{d t}\right|_{r=100}=400 \pi \mathrm{cm}^{2} / \mathrm{sec}\)

Exercise \(\PageIndex{15}\)

If water is being pumped into a spherical balloon at the rate of 100 grams per second, find the rate of change of the radius \(r\) of the balloon when the radius of the balloon is \(r=15\) centimeters.

As an important special case of the chain rule, suppose \(n \neq 0\) is an integer, \(g\) is a differentiable function, and \(h(x)=(g(x))^{n}\). Then \(h\) is the composition of \(f(x)=x^{n}\) with \(g,\) and so, using the chain rule, \[h^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)=n(g(x))^{n-1} g^{\prime}(x) .\] If we let \(u=g(x),\) we could also express this result as \[\frac{d}{d x} u^{n}=n u^{n-1} \frac{d u}{d x} .\]

Answer

\(\left.\frac{d r}{d t}\right|_{r=15}=\frac{1}{9 \pi}\) centimeters per second

Example \(\PageIndex{16}\)

With \(n=10\) and \(g(x)=x^{2}+3,\) we have

\[\frac{d}{d x}\left(x^{2}+3\right)^{10}=10\left(x^{2}+3\right)^{9}(2 x)=20 x\left(x^{2}+3\right)^{9} .\]

Example \(\PageIndex{17}\)

If

\[f(x)=\frac{15}{\left(x^{4}+5\right)^{2}} ,\] then we may apply the previous result with \(n=-2\) and \(g(x)=x^{4}+5\) to obtain \[f^{\prime}(x)=-30\left(x^{4}+5\right)^{-3}\left(4 x^{3}\right)=-\frac{120 x^{3}}{\left(x^{4}+5\right)^{3}} .\] We may use the previous result to derive yet another extension to the power rule. If \(n \neq 0\) is an integer and \(y=x^{\frac{1}{n}},\) then \(y^{n}=x,\) and so, assuming \(y\) is differentiable, \[\frac{d}{d x} y^{n}=\frac{d}{d x} x .\] Hence \[n y^{n-1} \frac{d y}{d x}=1 ,\] from which if follows that \[\frac{d y}{d x}=\frac{1}{n y^{n-1}}=\frac{1}{n} y^{1-n}=\frac{1}{n}\left(x^{\frac{1}{n}}\right)^{1-n}=\frac{1}{n} x^{\frac{1}{n}-1} ,\] showing that the power rule works for rational powers of the form \(\frac{1}{n} .\) Note that the above derivation is not complete since we began with the assumption that \(y=x^{\frac{1}{n}}\) is differentiable. Although it is beyond the scope of this text, it may be shown that this assumption is justified for \(x>0\) if \(n\) is even, and for all \(x \neq 0\) if \(n\) is odd. Now if \(m \neq 0\) is also an integer, we have, using the chain rule as above, \[\begin{aligned} \frac{d}{d x} x^{\frac{r m}{m}} &=\frac{d}{d x}\left(x^{\frac{1}{m}}\right)^{m} \\ &=m\left(x^{\frac{1}{n}}\right)^{m-1} \frac{1}{n} x^{\frac{1}{n}-1} \\ &=\frac{m}{n} x^{\frac{m-1}{n}+\frac{1}{n}-1} \\ &=\frac{m}{n} x^{\frac{m}{n}-1} .\end{aligned}\] Hence we now see that the power rule holds for any non-zero rational exponent.

Theorem \(\PageIndex{10}\)

If \(r \neq 0\) is any rational number, then

\[\frac{d}{d x} x^{r}=r x^{r-1} .\]

Example \(\PageIndex{18}\)

With \(r=\frac{1}{2}\) in the previous theorem, we have

\[\frac{d}{d x} \sqrt{x}=\frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}} ,\] in agreement with our earlier direct computation.

Example \(\PageIndex{19}\)

If \(y=x^{\frac{2}{3}},\) then

\[\frac{d y}{d x}=\frac{2}{3} x^{-\frac{1}{2}}=\frac{2}{3 x^{\frac{1}{3}}} .\] Note that \(\frac{d y}{d x}\) is not defined at \(x=0,\) in agreement with our earlier result showing that \(y\) is not differentiable at \(0 .\)

Exercise \(\PageIndex{16}\)

Find the derivative of \(f(x)=5 x^{\frac{4}{5}}\).

We may now generalize 1.7 .44 as follows: If \(u\) is a differentiable function of \(x\) and \(r \neq 0\) is a rational number, then \[\frac{d}{d x} u^{r}=r u^{r-1} \frac{d u}{d x} .\]

Answer

\(f^{\prime}(x)=\frac{4}{x^{\frac{1}{5}}}\)

Example \(\PageIndex{20}\)

If \(f(x)=\sqrt{x^{2}+1},\) then

\[f^{\prime}(x)=\frac{1}{2}\left(x^{2}+1\right)^{-\frac{1}{2}}(2 x)=\frac{x}{\sqrt{x^{2}+1}} .\]

Example \(\PageIndex{21}\)

If

\[g(t)=\frac{1}{t^{4}+5} ,\] then \[g^{\prime}(t)=(-1)\left(t^{4}+5\right)^{-2}\left(4 t^{3}\right)=-\frac{4 t^{3}}{\left(t^{4}+5\right)^{2}} .\]

Exercise \(\PageIndex{17}\)

Find the derivative of

\[y=\frac{4}{\sqrt{x^{2}+4}} .\]

Answer

\(\frac{d y}{d x}=-\frac{4 x}{\left(x^{2}+4\right)^{\frac{3}{2}}}\)

Exercise \(\PageIndex{18}\)

Find the derivative of \(f(x)=\left(x^{2}+3 x-5\right)^{10}\left(3 x^{4}-6 x+4\right)^{12}\).

Answer

\(\begin{aligned} f^{\prime}(x)=12\left(x^{2}+3 x-5\right)^{10} &\left(3 x^{4}-6 x+4\right)^{11}\left(12 x^{3}-6\right)+\\ & 10\left(x^{2}+3 x-5\right)^{9}\left(3 x^{4}-6 x+4\right)^{12}(2 x+3) \end{aligned}\)

Trigonometric Functions

If \(y=\sin (x)\) and \(w=\cos (x),\) then, for any infinitesimal \(d x\),

\[ \begin{aligned} d y &=\sin (x+d x)-\sin (x) \\ &=\sin (x) \cos (d x)+\sin (d x) \cos (x)-\sin (x) \\ &=\sin (x)(\cos (d x)-1)+\cos (x) \sin (d x) \end{aligned}\] and \[\begin{aligned} d w &=\cos (x+d x)-\cos (x) \\ &=\cos (x) \cos (d x)-\sin (x) \sin (d x)-\cos (x) \\ &=\cos (x)(\cos (d x)-1)-\sin (x) \sin (d x) . \end{aligned}\] Hence, if \(d x \neq 0\), \[\frac{d y}{d x}=\cos (x) \frac{\sin (d x)}{d x}-\sin (x) \frac{1-\cos (d x)}{d x}\] and \[\frac{d w}{d x}=-\sin (x) \frac{\sin (d x)}{d x}+\cos (x) \frac{1-\cos (d x)}{d x} .\] Now from \((1.5 .13)\) we know that \[0 \leq 1-\cos (d x) \leq \frac{(d x)^{2}}{2} ,\] and so \[0 \leq \frac{1-\cos (x)}{d x} \leq \frac{d x}{2} .\] Hence \[\frac{1-\cos (d x)}{d x}\] is an infinitesimal. Moreover, from \((1.5 .36),\) we know that \[\frac{\sin (d x)}{d x} \simeq 1 .\] Hence \[\frac{d y}{d x} \simeq \cos (x)(1)-\sin (x)(0)=\cos (x)\] and \[\frac{d w}{d x} \simeq-\sin (x)(1)+\cos (x)(0)=-\sin (x) .\] That is, we have shown the following.

Theorem \(\PageIndex{11}\)

For all real values \(x\),

\[\frac{d}{d x} \sin (x)=\cos (x)\] and \[\frac{d}{d x} \cos (x)=-\sin (x) .\]

Example \(\PageIndex{22}\)

Using the chain rule,

\[\frac{d}{d x} \cos (4 x)=-\sin (4 x) \frac{d}{d t}(4 x)=-4 \sin (4 x) .\]

Example \(\PageIndex{23}\)

If \(f(t)=\sin ^{2}(t),\) then, again using the chain rule,

\[f^{\prime}(t)=2 \sin (t) \frac{d}{d t} \sin (t)=2 \sin (t) \cos (t) .\]

Example \(\PageIndex{24}\)

If \(g(x)=\cos \left(x^{2}\right),\) then

\[g^{\prime}(x)=-\sin \left(x^{2}\right)(2 x)=-2 x \cos \left(x^{2}\right) .\]

Example \(\PageIndex{25}\)

If \(f(x)=\sin ^{3}(4 x),\) then, using the chain rule twice,

\[f^{\prime}(x)=3 \sin ^{2}(4 x) \frac{d}{d x} \sin (4 x)=12 \sin ^{2}(4 x) \cos (4 x) .\]

Exercise \(\PageIndex{19}\)

Find the derivatives of

\[y=\cos (3 t+6) \text { and } w=\sin ^{2}(t) \cos ^{2}(4 t) .\]

Answer

\(\frac{d y}{d t}=-3 \sin (3 t+6)\)

\(\frac{d w}{d t}=-8 \sin ^{2}(t) \cos (4 t) \sin (4 t)+2 \sin (t) \cos (t) \cos ^{2}(4 t)\)

Exercise \(\PageIndex{20}\)

Verify the following:

\[\begin{array}{ll}{\text { (a) } \frac{d}{d t} \tan (t)=\sec ^{2}(t)} & {\text { (b) } \frac{d}{d t} \cot (t)=-\csc ^{2}(t)} \\ {\text { (c) } \frac{d}{d t} \sec (t)=\sec (t) \tan (t)} & {\text { (d) } \frac{d}{d t} \csc (t)=-\csc (t) \cot (t)}\end{array}\]

Exercise \(\PageIndex{21}\)

Find the derivative of \(y=\sec ^{2}(3 t)\).

Answer

\(\frac{d y}{d x}=6 \sec ^{2}(3 t) \tan (3 t)\)

Exercise \(\PageIndex{22}\)

Find the derivative of \(f(t)=\tan ^{2}(3 t)\).

Answer

\(f^{\prime}(t)=6 \tan (3 t) \sec ^{2}(3 t)\)