How do you find the derivative of #(sinx^2)/(tanx)#?

2 Answers
Mar 3, 2018

Answer:

#(2xcos(x^2)tanx-sec^2xsin(x^2))/tan^2x#

Explanation:

According to the quotient rule, #(f/g)'=(f'g-fg')/g^2#

Here, #f=sin(x^2)# and #g=tan(x)#

#d/dx(sinx^2/tanx)=(d/dx(sin(x^2))*tanx-d/dx(tanx)*sin(x^2))/(tanx)^2#

We first compute:

#d/dx(sin(x^2))#

If we take the chain rule, where #f(u)=sin(x^2)#, we have #f=sin(u)# and #u=x^2#

So we have:

#d/(du)sin(u)*d/dxx^2#

#cosu*2x#

And as #u=x^2#, we have:

#d/dx(sin(x^2))=2xcos(x^2)#

Then ,we compute:

#d/dxtanx=sec^2(x)#

So we can input into our original equation:

#(2xcos(x^2)tanx-sec^2xsin(x^2))/tan^2x#

The answer.

Mar 3, 2018

Answer:

#2xcosx^2cotx-sinx^2csc^2x#

Explanation:

Let #f(x)=sinx^2#, and #g(x)=tanx#

The formula is

#(f/g)^'=(f^'g-g^'f)/g^2#

The derivative of #sinx# is #cosx#, but here we have #sinx^2# so the derivative will be #coscolor(red)(x^2# multiplied by the derivative of #color(red)(x^2# which is #2x#, so #2xcosx^2#

So #color(blue)(f^'=2xcosx^2#

#g^'# is the derivative of #tanx# which is #sec^2x#

So #color(green)(g^'=sec^2x#

Substituting these values in we get:-

#(f/g)^'=(color(blue)(2xcosx^2)tanx-color(green)(sec^2x)sinx^2)/color(brown)(tan^2x)#

Now we just simplify, separate the denominator:-

#(2xcosx^2)/tanx-(color(green)((1/cos^2x))sinx^2)/color(brown)(sin^2x/cos^2x)#

#=2xcosx^2cotx-sinx^2/cancel(cos^2x)*cancel(cos^2x)/sin^2x#

#=2xcosx^2cotx-sinx^2csc^2x#