Chapter 6.1 and 6.2 Practice
Questions
Compute the derivative of the following functions.
- \(f(x)=x^3-3^x\)
- \(f(x)=\sin(4)\cos(x)+5\)
- \(f(x)=e^x\tan(x)\)
- \(f(x)=\ln(x)+x^2\arctan(x)\)
- \(f(x)=\frac{\log_3(x)}{\log_2(x)}\)
- \(f(x)=\frac{\log_3(x)}{\log_2(5)}\)
- \(f(x)=-1.5e^{x}+\frac{1}{\sqrt{x}}-\sec(x)\)
- \(f(x)=g(x)e^x\), where \(g\) is a differentiable function (since you don’t have a formula for \(g\) just write down \(g'\) for derivative of \(g\))
- \(f(x)=\ln(x-1)\)
- \(f(x)=\frac{1}{4}e^{x^2}\)
- \(f(x)=\sqrt{\sin(x)}\)
- \(f(x)=\frac{1}{\sin^2(x)}\)
- \(f(x)=\log_2(4)(x^2-x)^2\)
- \(f(x)=\frac{\sin(x)}{(4x-1)^2}\)
- \(f(x)=e^{\tan(x)}\)
- \(f(x)=e^{g(x)}\), where \(g\) is a differentiable function
- \(f(x)=(\arctan(x))^n\), where \(n\) is a non-zero real number.
- \(f(x)=(g(x))^n\) where \(g\) is a differentiable function and \(n\) is a non-zero real number
- \(f(x)=10e^{x}-40\)
- \(f(x)=\pi^2\cos(2x)\)
- \(f(x)=\frac{40}{\sqrt{x+2}}\)
- \(f(x)=\frac{\arctan(x)}{\sin(x)+3}\)
- \(f(x)=\log_2(x)(x+4^x)^2\)
- \(f(x)=g(x)\ln(x)\), where \(g\) is a differentiable function
Find \(y'\) or \(\frac{dy}{dx}\), where \(y\) is defined implicitly by the equations given below:
- \((x-1)^2+xy+(y+3)^2=1\)
- \(\sin(x)y=e^y\)
- \(\sin(y)x=x+y\)
- \(y=x^{\tan(x)}\)
- \(y=(x+3)^{\sec(x)}e^x\)
For each function \(f\) and point \(p\), determine the equation of the line tangent to \(f\) at \(p\).
- \(f(x)=5x^3+20x+1\), \(p=(2,81)\)
- \(f(x)=4x\exp(x)\), \(p=(0,0)\)
- \(f(x)=\sqrt{x^2+1}\), \(p=(\sqrt{3},2)\)
- \(f(x)=\frac{x}{x+2}\), \(p=\left(6,\frac{3}{4}\right)\)
- \(f(x)=\frac{\cos(x)}{x+1}\), \(p=(0,1)\)
For each function \(f\) and \(x_0\) given, use the local linear approximation to the given value \(f(a)\).
- \(f(x)=\frac{3}{x^\frac{1}{4}}\), \(x_0=7\) and \(f(7.5)\)
- \(f(x)=\cot(x)\), \(x_0=1\) and \(f(1.05)\)
- \(f(x)=6^x\), \(x_0=5\) and \(f(5.5)\)
Use Newton’s method to construct a sequence \((x_n)\) to approximate the value \(6^\frac{1}{3}\). Take \(x_1=2\) and write down the values of \(x_2\) and \(x_3\).
Take \(f\) to be the function that is given by \(f(x,y)=4x^5y^6+10y^3\). Determine \(f_x(x,y)\), \(f_y(x,y)\), \(f_x(1,2)\) and \(f_y(1,2)\).
The equation \[x^4y-xy^5=-11200\] implicitly defines \(y\) as a function of \(x\) in an open rectangle around the point \((4,5).\) Determine an equation for the line that is tangent to the solution set to the equation at the point \((4,5)\).
Take \(f\) to be the function that is given by \[f(x)=(4x+3)^9(\tan(x)+8)^6(8x+2)^5(\sin(x)+4)^5.\] Use logarithmic differentiation to differentiate the function \(f\).
A spherical balloon is being inflated with air at a rate so its volume is increasing at a rate of \(200\) centimeters cubed per second. Determine the rate of change of the radius of the balloon when the diameter is \(80\) centimeters.
There are two cars, Car A and Car B, that are headed towards the same intersection. Car A travels east at \(65\) miles per hour and Car B travels south at \(55\) miles per hour. Determine the rate at which the cars approach each other when Car A is \(0.5\) miles from the intersection and Car B is \(1.2\) miles from the intersection.
A \(15\) feet long ladder, that is initially resting against a vertical wall, begins to slide. The bottom of the ladder slides from the wall at a rate of \(2\) feet per second. Determine how fast the top of the ladder is sliding down when the bottom of the ladder is \(9\) feet fro the wall.
Answers
- \(f'(x)=3x^2-3^x\ln(3)\)
- \(f'(x)=-\sin(4)\sin(x)\)
- \(f'(x)=e^x\tan(x)+e^x\sec^2(x)\)
- \(f(x)=\frac{1}{x}+2x\arctan(x)+\frac{x^2}{x^2+1}\)
- \(f'(x)=\frac{\frac{1}{x\ln(3)}\log_2(x)-\log_3(x)\cdot \frac{1}{x\ln(2)}}{(\log_2(x))^2}\)
- \(f'(x)=\frac{1}{\log_2(5)\ln(3)x}\)
- \(f'(x)=-1.5e^{x}-\frac{1}{2x^{3/2}}-\sec(x)\tan(x)\)
- \(f'(x)=g'(x)e^x+g(x)e^x\)
- \(f'(x)=\frac{1}{x-1}\)
- \(f'(x)=\frac{1}{2}xe^{x^2}\)
- \(f'(x)=\frac{1}{2}\cos(x)(\sin(x))^{-1/2}\) or \(\frac{\cos(x)}{2\sqrt{\sin(x)}}\)
- \(f'(x)=-2\cos(x)(\sin(x))^{-3}\) or \(\frac{2\cos(x)}{\sin^3(x)}\)
- \(f'(x)=4(x^2-x) (2x-1)\)
- \(f'(x)=\frac{\cos(x)(4x-1)^2-8\sin(x)(4x-1)}{(4x-1)^4}\)
- \(f'(x)=e^{\tan(x)}\sec^2(x)\)
- \(f'(x)=e^{g(x)}g'(x)\)
- \(f'(x)=n(\arctan(x))^{n-1}\frac{1}{x^2+1}\)
- \(f(x)=n(g(x))^{n-1}g'(x)\)
- \(f'(x)=10e^{x}\)
- \(f'(x)=-2\pi^2\sin(2x)\)
- \(f'(x)=-\frac{40}{2(x+2)^{3/2}}\)
- \(f'(x)=\frac{\frac{\sin(x)+3}{x^2+1}-\arctan(x)\cos(x)}{(\sin(x)+3)^2}\)
- \(f'(x)=\frac{(x+4^x)^2}{x\ln(2)}+2\log_2(x)(x+4^x)(1+4^x\ln(4))\)
- \(f'(x)=g'(x)\ln(x)+\frac{g(x)}{x}\)
- \(y'=\frac{-2x+2-y}{x+2y+6}\)
- \(y'=\frac{y\cos(x)}{\sin(x)-e^y}\)
- \(y'=\frac{1-\sin(y)}{x\cos(y)-1}\)
- \(y'=x^{\tan(x)}\left(\sec^2(x)\ln(x)+\frac{\tan(x)}{x}\right)\)
- \(y'=(x+3)^{\sec(x)}e^x\left(\sec(x)\tan(x)\ln(x+3)+\frac{\sec(x)}{x+3}+1\right)\)
- \(L(x)=80(x-2)+81\)
- \(L(x)=4x\)
- \(L(x)=\frac{\sqrt{3}}{2}\left(x-\sqrt{3}\right)+2\)
- \(L(x)=\frac{1}{32}(x-6)+\frac{3}{4}\)
- \(L(x)=-x+1\)
- \(f(7.5)\approx 1.81142\)
- \(f(1.05)\approx 0.571478\)
- \(f(5.5)\approx 14742.360816\)
\(x_2=1.8\bar{3}\) and \(x_3=1.817265\dots\)
\(f_x(x,y)=20x^4y^6\), \(f_y(x,y)=24x^5y^5+30y^2\), \(f_x(1,2)=1280\) and \(f_y(1,2)=888\).
\(L(x)=-\frac{1845}{12244}(x-4)+5\)
\(f'(x)=(4x+3)^9(\tan(x)+8)^6(8x+2)^5(\sin(x)+4)^5\left(\frac{36}{4x+3}+\frac{6\sec^2(x)}{\tan(x)+8}+\frac{40}{8x+2}+\frac{5\cos(x)}{\sin(x)+4}\right)\)
\(0.01\) centimeters per second
Approximately \(-75.77\) miles per hour.
\(-1.5\) feet per second.