Exercise
Find the derivative of the following function:
f(x)=\frac{x{(2x-3)}^5}{e^{3x^2+1}}
Final Answer
Solution
f(x)=\frac{x{(2x-3)}^5}{e^{3x^2+1}}
Using Derivative formulas and the quotient rule in Derivative Rules, we get the derivative:
f'(x)=\frac{[{(2x-3)}^5+5x{(2x-3)}^4\cdot 2]\cdot e^{3x^2+1}-x{(2x-3)}^5\cdot 6x\cdot e^{3x^2+1}}{{(e^{3x^2+1})}^2}=
One can simplify the derivative:
=\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot [(2x-3)+10x-6x^2(2x-3)]=
=\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (2x-3+10x-12x^3+18x^2)=
=\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (-12x^3+18x^2+12x-3)
Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!