Exercise:
For the following exercises, given \(y=f(u)\) and \(u=g(x)\), find dydx by using Leibniz’s notation for the chain rule: \(\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.\)
214) \(y=3u−6,u=2x^2\)
215) \(y=6u^3,u=7x−4\)
Solution: \(18u^2⋅7=18(7x−4)^2⋅7\)
216) \(y=sinu,u=5x−1\)
217) \(y=cosu,u=\frac{−x}{8}\)
Solution: \(−sinu⋅\frac{−1}{8=}−sin(\frac{−x}{8})⋅\frac{−1}{8}\)
218) \(y=tanu,u=9x+2\)
219) \(y=\sqrt{4u+3},u=x^2−6x\)
Solution: \(\frac{8x−24}{2\sqrt{4u+3}}=\frac{4x−12}{\sqrt{4x^2−24x+3}}\)
For each of the following exercises,
a. decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and
b. find \(\frac{dy}{dx}\) as a function of \(x\).
220) \(y=(3x−2)^6\)
221) \(y=(3x^2+1)^3\)
Solution: a. \(u=3x^2+1\); b. \(18x(3x^2+1)^2\)
222) \(y=sin^5(x)\)
For each of the following exercises, find \(\frac{dy}{dx}\) as a function of \(x\).
223) \(y=(\frac{x}{7}+\frac{7}{x})^7\)
Solution: \(a. f(u)=u^7,u=\frac{x}{7}+\frac{7}{x}; b. 7(\frac{x}{7}+\frac{7}{x})^6⋅(\frac{1}{7}−\frac{7}{x^2})\)
224) \(y=tan(secx)\)
225) \(y=csc(πx+1)\)
Solution: \(a. f(u)=cscu,u=πx+1; b. −πcsc(πx+1)⋅cot(πx+1)\)
226) \(y=cot^2x\)
227) \(y=−6sin^{−3}x\)
a. \(f(u)=−6u^{−3},u=sinx, b. 18sin^{−4}x⋅cosx\)
For the following exercises, find \(\frac{dy}{dx}\) for each function.
228) \(y=(3x^2+3x−1)^4\)
229) \(y=(5−2x)^{−2}\)
Solution: \(\frac{4}{(5−2x)^3}\)
230) \(y=cos^3(πx)\)
231) \(y=(2x^3−x^2+6x+1)^3\)
Solution: \(6(2x^3−x^2+6x+1)^2(3x^2−x+3)\)
232) \(y=\frac{1}{sin^2(x)}\)
233) \(y=(tanx+sinx)^{−3}\)
Soution: \(−3(tanx+sinx)^{−4}⋅(sec^2x+cosx)\)
234) \(y=x^2cos^4x\)
235) \(y=sin(cos7x)\)
Solution: \(−7cos(cos7x)⋅sin7x\)
236) \(y=\sqrt{6+secπx^2}\)
237) \(y=cot^3(4x+1)\)
Solution: \(−12cot^2(4x+1)⋅csc^2(4x+1)\)
238) Let \(y=[f(x)]^3\) and suppose that \(f′(1)=4\) and \(\frac{dy}{dx}=10\) for \(x=1\). Find \(f(1)\).
239) Let \(y=(f(x)+5x^2)^4\) and suppose that \(f(−1)=−4\) and \(\frac{dy}{dx}=3\) when \(x=−1\). Find \(f′(−1)\)
Solution: \(10\frac{3}{4}\)
240) Let \(y=(f(u)+3x)^2\) and \(u=x^3−2x\). If \(f(4)=6\) and \(\frac{dy}{dx}=18\) when \(x=2\), find \(f′(4)\).
241) [T] Find the equation of the tangent line to \(y=−sin(\frac{x}{2})\) at the origin. Use a calculator to graph the function and the tangent line together.
Solution: \(y=\frac{−1}{2}x\)
242) [T] Find the equation of the tangent line to \(y=(3x+\frac{1}{x})^2\) at the point \((1,16)\). Use a calculator to graph the function and the tangent line together.
243) Find the \(x\) -coordinates at which the tangent line to \(y=(x−\frac{6}{x})^8\) is horizontal.
Solution: \(x=±\sqrt{6}\)
244) [T] Find an equation of the line that is normal to \(g(θ)=sin2^(πθ)\) at the point \((\frac{1}{4},\frac{1}{2})\). Use a calculator to graph the function and the normal line together.
For the following exercises, use the information in the following table to find \(h′(a)\) at the given value for \(a\).
| \(x\) |
\(f(x)\) |
\(f'(x)\) |
\(g(x)\) |
\(g'(x)\) |
| 0 |
2 |
5 |
0 |
2 |
| 1 |
1 |
−2 |
3 |
0 |
| 2 |
4 |
4 |
1 |
−1 |
| 3 |
3 |
−3 |
2 |
3 |
245) \(h(x)=f(g(x));a=0\)
Solution: \(10\)
246) \(h(x)=g(f(x));a=0\)
247) \(h(x)=(x^4+g(x))^{−2};a=1\)
Solution: \(−\frac{1}{8}\)
248) \(h(x)=(\frac{f(x)}{g(x)})^2;a=3\)
249) \(h(x)=f(x+f(x));a=1\)
Solution: \(−4\)
250) \(h(x)=(1+g(x))^3;a=2\)
251) \(h(x)=g(2+f(x^2));a=1\)
Solution: \(−12\)
252) h(x)=f(g(sinx));a=0
253) [T] The position function of a freight train is given by
\(s(t)=100(t+1)^{−2}\), with \(s\) in meters and \(t\) in seconds. At time \(t=6\)s, find the train’s
a. velocity and
b. acceleration.
c. Using a. and b. is the train speeding up or slowing down?
Solution: \(a. −\frac{200}{343}\) m/s, b. \(\frac{600}{2401}\) m/s^2, c. The train is slowing down since velocity and acceleration have opposite signs.
254) [T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t is measured in seconds and \(s\) is in inches:
\(s(t)=−3cos(πt+\frac{π}{4}).\)
a. Determine the position of the spring at \(t=1.5\) s.
b. Find the velocity of the spring at \(t=1.5\) s.
255) [T] The total cost to produce \(x\) boxes of Thin Mint Girl Scout cookies is \(C\) dollars, where \(C=0.0001x^3−0.02x^2+3x+300.\) In \(t\) weeks production is estimated to be \(x=1600+100t\) boxes.
a. Find the marginal cost \(C′(x).\)
b. Use Leibniz’s notation for the chain rule, \(\frac{dC}{dt}=\frac{dC}{dx}⋅\frac{dx}{dt}\), to find the rate with respect to time \(t\) that the cost is changing.
c. Use b. to determine how fast costs are increasing when \(t=2\) weeks. Include units with the answer.
Solution: \(a. C′(x)=0.0003x^2−0.04x+3\)
\(b. dCdt=100⋅(0.0003x^2−0.04x+3)\) c. Approximately $90,300 per week
256) [T] The formula for the area of a circle is \(A=πr^2\), where \(r\) is the radius of the circle. Suppose a circle is expanding, meaning that both the area \(A\) and the radius \(r\) (in inches) are expanding.
a. Suppose \(r=2−\frac{100}{(t+7)^2}\) where \(t\) is time in seconds. Use the chain rule \(\frac{dA}{dt}=\frac{dA}{dr}⋅\frac{dr}{dt}\) to find the rate at which the area is expanding.
b. Use a. to find the rate at which the area is expanding at \(t=4\) s.
257) [T] The formula for the volume of a sphere is \(S=\frac{4}{3}πr^3\), where \(r\) (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.
a. Suppose \(r=\frac{1}{(t+1)^2}−\frac{1}{12}\) where t is time in minutes. Use the chain rule \(\frac{dS}{dt}=\frac{dS}{dr}⋅\frac{dr}{dt}\) to find the rate at which the snowball is melting.
b. Use a. to find the rate at which the volume is changing at \(t=1\) min.
Solution: \(a. \frac{dS}{dt}=−\frac{8πr^2}{(t+1)^3}\) b. The volume is decreasing at a rate of \(−\frac{π}{36}\) \(ft^3\)/min
258) [T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function \(T(x)=94−10cos[\frac{π}{12}(x−2)]\), where \(x\) is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.
259) [T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function \(D(t)=5sin(\frac{π}{6}t−\frac{7π}{6})+8\), where \(t\) is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.
Solution: \(~2.3\) ft/hr
