Exercise
Find the derivative of the following function:
f ( x ) = x ( 2 x − 3 ) 5 e 3 x 2 + 1 f(x)=\frac{x{(2x-3)}^5}{e^{3x^2+1}} f ( x ) = e 3 x 2 + 1 x ( 2 x − 3 ) 5
Final Answer
Show final answer
f'(x)=\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (-12x^3+18x^2+12x-3)
Solution
f ( x ) = x ( 2 x − 3 ) 5 e 3 x 2 + 1 f(x)=\frac{x{(2x-3)}^5}{e^{3x^2+1}} f ( x ) = e 3 x 2 + 1 x ( 2 x − 3 ) 5
Using Derivative formulas Derivative Rules
f ′ ( x ) = [ ( 2 x − 3 ) 5 + 5 x ( 2 x − 3 ) 4 ⋅ 2 ] ⋅ e 3 x 2 + 1 − x ( 2 x − 3 ) 5 ⋅ 6 x ⋅ e 3 x 2 + 1 ( e 3 x 2 + 1 ) 2 = 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}= f ′ ( x ) = ( e 3 x 2 + 1 ) 2 [ ( 2 x − 3 ) 5 + 5 x ( 2 x − 3 ) 4 ⋅ 2 ] ⋅ e 3 x 2 + 1 − x ( 2 x − 3 ) 5 ⋅ 6 x ⋅ e 3 x 2 + 1 =
One can simplify the derivative:
= ( 2 x − 3 ) 4 e 3 x 2 + 1 ⋅ [ ( 2 x − 3 ) + 10 x − 6 x 2 ( 2 x − 3 ) ] = =\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot [(2x-3)+10x-6x^2(2x-3)]= = e 3 x 2 + 1 ( 2 x − 3 ) 4 ⋅ [ ( 2 x − 3 ) + 1 0 x − 6 x 2 ( 2 x − 3 ) ] =
= ( 2 x − 3 ) 4 e 3 x 2 + 1 ⋅ ( 2 x − 3 + 10 x − 12 x 3 + 18 x 2 ) = =\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (2x-3+10x-12x^3+18x^2)= = e 3 x 2 + 1 ( 2 x − 3 ) 4 ⋅ ( 2 x − 3 + 1 0 x − 1 2 x 3 + 1 8 x 2 ) =
= ( 2 x − 3 ) 4 e 3 x 2 + 1 ⋅ ( − 12 x 3 + 18 x 2 + 12 x − 3 ) =\frac{{(2x-3)}^4}{e^{3x^2+1}}\cdot (-12x^3+18x^2+12x-3) = e 3 x 2 + 1 ( 2 x − 3 ) 4 ⋅ ( − 1 2 x 3 + 1 8 x 2 + 1 2 x − 3 )
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