Question #217eb

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Question #217eb

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Answer 1 · 2016-11-30T14:20:27

$f'(x) = (sin^2x + sinx + xcosx)/(1 + sinx)^2$

Explanation:

First of all, the function can be simplified.

Call the function $f(x)$ .

$f(x) = (xtanx)/(1/cosx + sinx/cosx)$

$f(x) = (xtanx)/((1 + sinx)/cosx)$

$f(x) = (xtanx(cosx))/(1 + sinx)$

$f(x) = (xsinx/cosx(cosx))/(1 + sinx)$

$f(x) = (xsinx)/(1 + sinx)$

We now differentiate this using the quotient rule. The quotient rule states that for a function $color(red)(f(x) = (g(x))/(h(x))$ , the derivative is given by $color(red)(f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2$ . We know the derivative of the numerator is $sinx + xcosx$ (by the product rule), and that the derivative of the denominator is $cosx$ .

Hence:

$f'(x) = ((sinx + xcosx)(1 + sinx) - xsinxcosx)/(1 + sinx)^2$

$f'(x) = (sinx + xcosx+ sin^2x + xcosxsinx - xcosxsinx)/(1 + sinx)^2$

$f'(x) = (sin^2x + sinx + xcosx)/(1 + sinx)^2$

Hopefully this helps!

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Question #217eb

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