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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