How do you differentiate y=sinx+x^2tan^-1x?

2 Answers
Sonnhard
Jun 13, 2018

y'(x)=cos(x)+2xarctan(x)+x^2/(1+x^2)

Explanation:

Note that
(arctan(x))'=1/(1+x^2)
and by the sum and product rule we get

y'(x)=cos(x)+2xarctan(x)+x^2/(1+x^2)

Jim G.
Jun 13, 2018

dy/dx=cosx+x^2/(1+x^2)+2xtan^-1x

Explanation:

"differentiate "x^2tan^-1x" using the "color(blue)"product rule"

"given "y=f(x)g(x)" then"

dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"

f(x)=x^2rArrf'(x)=2x

g(x)=tan^-1xrArrg'(x)=1/(1+x^2)

d/dx(x^2tan^-1x)

=x^2/(1+x^2)+2xtan^-1x

dy/dx=cosx+x^2/(1+x^2)+2xtan^-1x

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