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Calculating Derivative – A quotient of a multiplication of polynoms and an exponential function – Exercise 6277

Exercise

Find the derivative of the following function:

f(x)=\frac{x{(2x-3)}^5}{e^{3x^2+1}}

Final Answer


f'(x)=\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (-12x^3+18x^2+12x-3)

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)

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