How do you find the derivative of $f(x) = tan^2(x)$ ?

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Answer 1 · 2016-07-14T02:14:23

$f(x) = sin^2x/cos^2x$

Let $f(x) = g(x)/(h(x))$ , so that $g(x) = sin^2x$ and $h(x) = cos^2x$ . Then $f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2$ .

Let's differentiate $g(x)$ and $h(x)$ .

$g(x) = (sinx)(sinx)$

$g'(x) = cosxsinx + cosxsinx$

$g'(x) = 2cosxsinx$

$g'(x) = sin2x$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$h'(x) = (cosx)(cosx)$

$h'(x) = -sinxcosx - sinxcosx$

$h'(x) = -2sinxcosx$

$h'(x) = -(2sinxcosx)$

$h'(x) = -sin2x$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2$ .

$f'(x) = ((sin2x xx cos^2x) - (-sin2x xx sin^2x))/(cos^2x)^2$

$f'(x) = ((2sinxcosx xx cos^2x) - (-2sinxcosx xx sin^2x))/(cos^2x)^2$

$f'(x) = ((2sinxcos^3x) - (-2sin^3xcosx))/(cos^4x)$

$f'(x) = (2sinxcos^3x + 2sin^3xcosx)/(cos^4x)$

$f'(x) = (2sinxcosx(cos^2x + sin^2x))/cos^4x$

$f'(x) = (2sinx)/cos^3x$

$f'(x) = 2cotxsec^2x$

Hopefully this helps!

How do you find the derivative of f(x) = tan^2(x)f(x) = tan^2(x)?