Question

differentiation classwork problems
differentiate each of the following functions using the appropriate rules. write your solutions to five of the following problems on a full sheet of paper to turn in at the end of class. you do not need to simplify your answers.

1. $y = 3x^{6}-5sqrt3{x}+9x - 8$
2. $y = x^{2}sin(pi x)$
3. $f(t)=\frac{1 - t + 4t^{2}+6t^{4}}{t^{2}}$
4. $f(\theta)=\frac{cot\theta}{1+csc\theta}$
5. $g(x)=e^{2x}-\tan^{-1}(5x)$
6. $y = e^{xsec x}+e^{sqrt{pi}}$
7. $f(z)=(2z + 1)^{3}-ln(5^{z})$
8. $y=\frac{3^{x}}{cos^{5}x}$
9. $y=ln(\frac{sqrt3{t}}{1 + t^{2}})$
10. $y = 5e^{x}sqrt{7x^{4}+1}$
11. $y=cot(csc(5x))$
12. $h(t)=t^{2}e^{t^{5}}$
13. $y=\tan^{3}(x^{2}+1)$
14. $y=\tan^{-1}(xe^{x})$

extra problems!
ex1: find the equation of the tangent line to the graph of the function $f(x)=2x^{2}+3x - 6$ at $x = 1$.
ex2: find where the tangent line to the graph of $f(x)=sqrt{x}$ is parallel to the line $y = 3x - 2$.
ex3: find the $n$th derivative of the function $f(x)=e^{3x}$.
ex4: use the quotient rule to prove the following differentiation rule: $\frac{d}{dx}(sec x)=sec x\tan x$.
ex5: find the derivative $\frac{dy}{dx}$ using implicit differentiation: $5x^{5}-2xy^{2}-e^{y}=27y$

Explanation:

Step1: Recall power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $y = 3x^6-5\sqrt[3]{x}+9x - 8=3x^6-5x^{\frac{1}{3}}+9x - 8$.
$y^\prime=\frac{d}{dx}(3x^6)-\frac{d}{dx}(5x^{\frac{1}{3}})+\frac{d}{dx}(9x)-\frac{d}{dx}(8)$
$y^\prime = 3\times6x^{6 - 1}-5\times\frac{1}{3}x^{\frac{1}{3}-1}+9\times1-0$
$y^\prime = 18x^{5}-\frac{5}{3}x^{-\frac{2}{3}}+9$

Step2: Recall product - rule for differentiation

The product - rule states that if $y = u\cdot v$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v+uv^\prime$. For $y = x^{2}\sin(\pi x)$, let $u = x^{2}$ and $v=\sin(\pi x)$.
$u^\prime = 2x$ and $v^\prime=\pi\cos(\pi x)$
$y^\prime=2x\sin(\pi x)+x^{2}\pi\cos(\pi x)$

Step3: Simplify rational function and then differentiate

For $f(t)=\frac{1 - t + 4t^{2}+6t^{4}}{t^{2}}=t^{-2}-t^{-1}+4 + 6t^{2}$
$f^\prime(t)=\frac{d}{dt}(t^{-2})-\frac{d}{dt}(t^{-1})+\frac{d}{dt}(4)+\frac{d}{dt}(6t^{2})$
$f^\prime(t)=-2t^{-3}+t^{-2}+0 + 12t=-\frac{2}{t^{3}}+\frac{1}{t^{2}}+12t$

Step4: Use quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $f(\theta)=\frac{\cot\theta}{1+\csc\theta}$, let $u = \cot\theta=-\csc^{2}\theta$ and $v = 1+\csc\theta$, $v^\prime=-\csc\theta\cot\theta$
$f^\prime(\theta)=\frac{-\csc^{2}\theta(1 + \csc\theta)-\cot\theta(-\csc\theta\cot\theta)}{(1+\csc\theta)^{2}}$
$=\frac{-\csc^{2}\theta-\csc^{3}\theta+\csc\theta\cot^{2}\theta}{(1+\csc\theta)^{2}}$

Step5: Differentiate sum of exponential and inverse - trigonometric function

If $g(x)=e^{2x}-\tan^{-1}(5x)$
Using the chain - rule, $\frac{d}{dx}(e^{2x}) = 2e^{2x}$ and $\frac{d}{dx}(\tan^{-1}(5x))=\frac{5}{1 + 25x^{2}}$
$g^\prime(x)=2e^{2x}-\frac{5}{1 + 25x^{2}}$

Answer:

1. $y^\prime = 18x^{5}-\frac{5}{3}x^{-\frac{2}{3}}+9$
2. $y^\prime=2x\sin(\pi x)+x^{2}\pi\cos(\pi x)$
3. $f^\prime(t)=-\frac{2}{t^{3}}+\frac{1}{t^{2}}+12t$
4. $f^\prime(\theta)=\frac{-\csc^{2}\theta-\csc^{3}\theta+\csc\theta\cot^{2}\theta}{(1+\csc\theta)^{2}}$
5. $g^\prime(x)=2e^{2x}-\frac{5}{1 + 25x^{2}}$