Question

I need help with some Calculus Problems. I don't want the answer, I just want to know what steps I need to take to solve the problem.


1. If f(3)=$\frac{5pi}{3}$ and f'(3)=$\sqrt{3}$, find the equation of the line tangent to g(x)= cos(f(x)) at x=3.


2. If f(x)= $\sqrt{tan(2x+\frac{3pi}{4})}$, find $lim_{x-\ \textgreater \ [tex]\frac{pi}{4}$}[/tex] (f(x)-f($\frac{pi}{4}$))/(x-$\frac{pi}{4}$)


Thank you.

Answer 1

1. we need to find g(3) and g'(3) to write point slope form of the tangent line

g(3)=cos(f(3))=$cos(\frac{5 \pi}{3})=\frac{1}{2}$

now to find g'(3)

use chain rule

$g'(x)=\frac{d}{dx}cos(f(x))=-sin(f(x))*f'(x)$ so

$g'(3)=-sin(f(3))*f'(3)=-sin(\frac{5 \pi}{3})*\sqrt{3}=$ $(\frac{\sqrt{3}}{2})(\sqrt{3})=\frac{3}{2}$

the equation of a line that passes through (x1,y1) and having a slope of m is

$y-y_1=m(x-x_1)$

we know it passes through (3,1/2) and has a slope of 3/2

so the equation of the tangent line at x=3 is

$y-\frac{1}{2}=\frac{3}{2}(x-3)$

2.ah, difference quotient

$\lim_{x \to \frac{\pi}{4}} \frac{f(x)-f(\frac{\pi}{4})}{x-\frac{\pi}{4}}=$

$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{tan(2x+\frac{3 \pi}{4})}-\sqrt{tan(2(\frac{\pi}{4})+\frac{3 \pi}{4})}}{x-\frac{\pi}{4}}=$

$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{tan(2x+\frac{3 \pi}{4})}-1}{x-\frac{\pi}{4}}=$

I don't know how to simplify further