How do you differentiate f(x)= tanx twice using the quotient rule?

1 Answer
Dec 23, 2015

dy/dx=sec^2x

(d^2y)/dx^2=2sec^2xtanx

Explanation:

In order to start properly, let's just remember that tanx=(sinx)/(cosx) and now differentiate this quotient using the proper rule.

The quotient rule states that for a function y=f(x)/g(x), (dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2

So, as for your function:

f(x)=sinx
g(x)=cosx
f'(x)=cosx
g'(x)=-sinx

(dy)/(dx)=(cosxcosx-sinx(-sinx))/(cosx)^2=(cos^2x+sin^2x)/cos^2x

From trigonometric identities, we know that sin^2x+cos^2x=1

(dy)/(dx)=1/(cos^2x)=sec^2x

To find the second derivative, use the chain rule, which states that for a function y=f(g(x)), dy/dx=f'(g(x))g'(x).

First, rewrite dy/dx=cos^-2x

Then, according to the chain rule:

(d^2y)/dx^2=-2cos^-3xd/dx(cosx)

=>(-2(-sinx))/cos^3x

Your interpretation of a final answer could vary.

(d^2y)/dx^2=(2sinx)/cos^3x "or" (d^2y)/dx^2=2sec^2xtanx