Question #217eb

1 Answer
Noah G
Nov 30, 2016

#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!

Related questions
See all questions in Proving Identities
Impact of this question
1208 views around the world
You can reuse this answer
Creative Commons License