Exercise
Find the derivative of the following function:
f ( x ) = x 2 ⋅ e 3 x ⋅ ln ( 2 x ) f(x)=x^2\cdot e^{3x}\cdot \ln(2x) f ( x ) = x 2 ⋅ e 3 x ⋅ ln ( 2 x )
Final Answer
Show final answer
f'(x)=xe^{3x}(2\ln(2x)+3x\ln(2x)+1)
Solution
We simplify the function before differentiating:
f ( x ) = x 2 ⋅ e 3 x ⋅ ln ( 2 x ) = f(x)=x^2\cdot e^{3x}\cdot \ln(2x)= f ( x ) = x 2 ⋅ e 3 x ⋅ ln ( 2 x ) =
= ( x 2 ⋅ e 3 x ) ⋅ ln ( 2 x ) =(x^2\cdot e^{3x})\cdot \ln(2x) = ( x 2 ⋅ e 3 x ) ⋅ ln ( 2 x )
Using Derivative formulas Derivative Rules
f ′ ( x ) = ( 2 x e 3 x + x 2 ⋅ 3 e 3 x ) ln ( 2 x ) + ( x 2 e 3 x ⋅ 1 2 x ⋅ 2 = f'(x)=(2xe^{3x}+x^2\cdot 3e^{3x})\ln (2x)+(x^2e^{3x}\cdot\frac{1}{2x}\cdot 2= f ′ ( x ) = ( 2 x e 3 x + x 2 ⋅ 3 e 3 x ) ln ( 2 x ) + ( x 2 e 3 x ⋅ 2 x 1 ⋅ 2 =
One can simplify the derivative:
= 2 x e 3 x ln ( 2 x ) + 3 x 2 e 3 x ln ( 2 x ) + x e 3 x = =2xe^{3x}\ln(2x)+3x^2e^{3x}\ln(2x)+xe^{3x}= = 2 x e 3 x ln ( 2 x ) + 3 x 2 e 3 x ln ( 2 x ) + x e 3 x =
= x e 3 x ( 2 ln ( 2 x ) + 3 x ln ( 2 x ) + 1 ) =xe^{3x}(2\ln(2x)+3x\ln(2x)+1) = x e 3 x ( 2 ln ( 2 x ) + 3 x ln ( 2 x ) + 1 )
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