Find the equation of the tangent line to the graph of f at x=(1/2). How do I solve?
the function is f(x)=cos(3cos^(-1)x). I got the derivative to equal f'(x)= -sin(3cos^(-1)x)(-3/(√1-x^(2))

Question

Find the equation of the tangent line to the graph of f at x=(1/2). How do I solve? 
the function is f(x)=cos(3cos^(-1)x). I got the derivative to equal f'(x)= -sin(3cos^(-1)x)(-3/(√1-x^(2))

myininaya · Answer

I like to first write the given as an algebraic expression first.

myininaya · Answer

Also my thingy messed up so I will have to retype how to do it. :p

anonymous · Answer

im sorry! :/ thank you so much though!

myininaya · Answer

$$\text{ Let  } u=\arccos(x) \ \text{ So } \cos(u)= x=\frac{x}{1} =\frac{adjacent }{hyp}$$|dw:1413149993980:dw|
$$\cos(3 \arccos(x))=\cos(3 u)=\cos(2u+u)=\cos(2u)\cos(u)-\sin(2u)\sin(u)\ =[\cos^2(u)-\sin^2(u)]\cos(u)-2 \sin(u)\cos(u)\sin(u) \ = [x^2-(1-x^2)]x-2\sqrt{1-x^2}x \sqrt{1-x^2} \ =(2x^2-1)x-2x(1-x^2) \ =2x^3-x-2x+2x^3 \ =4x^3-3x $$
So you actually are given y=4x^3-3x