Question

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

Answer 1

`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`