How do you differentiate f(x)=(e^x-3x)(cotx+x^2) using the product rule?

1 Answer
Mar 1, 2016

\frac{d}{dx}((e^x-3x)(\cot (x)+x^2))=(e^x-3)(\cot (x)+x^2)+(-\frac{1}{\sin ^2(x)}+2x)(e^x-3x)

Explanation:

\frac{d}{dx}((e^x-3x)(\cot (x)+x^2))

Applying product rule as : (f\cdot g)^'=f^'\cdot g+f\cdot g^'

f=e^x-3x,g=\cot (x)+x^2

=\frac{d}{dx}(e^x-3x)(\cot (x)+x^2)+\frac{d}{dx}(\cot \(x)+x^2)(e^x-3x)............ eq^n (i)

Here,
\frac{d}{dx}(e^x-3x)=e^x-3
{Applying sum and differnce rule as: (f\pm g)^'=f^'\pm g^'

=\frac{d}{dx}(e^x)-\frac{d}{dx}(3x) and

\frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(3x)=3}

=e^x-3............. eq^n (ii)

Again,

\frac{d}{dx}(\cot (x)+x^2)=-\frac{1}{\sin ^2(x)}+2x

{Applying sum and differnce rule as: (f\pm g)^'=f^'\pm g^'
=\frac{d}{dx}(\cot (x))+\frac{d}{dx}(x^2) and

and we know the common derivative,

\frac{d}{dx}(\cot (x))=-\frac{1}{\sin ^2(x)} and also we know,#\frac{d}

{dx}(x^2)=2x#}

=-\frac{1}{\sin ^2(x)}+2x ............. eq^n (iii)

Finally,from eq^n (i), (ii) and (iii), we get,

=(e^x-3)(\cot (x)+x^2)+(-\frac{1}{\sin ^2(x)}+2x)(e^x-3x)