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Question

Mathematics Question on Exponential and Logarithmic Functions

The value of ddx[xnlog⁡axex]=\frac{d}{dx} [x^n \log_a xe^x ] =

A

exlog⁡ax+xn−1log⁡eae^x \log_a x + \frac{x^{n-1}}{\log_e a}

B

exxn−1[xlog⁡ax+1log⁡ea+xlog⁡ax] e^{x} x^{n-1} \left[x \log_{a} x + \frac{1}{\log_{e}a} + x \log_{a}x \right]

C

nxn−1log⁡a(xex)nx^{n-1} \log_{a} \left(xe^{x}\right)

D

xnlog⁡a(xex)x^{n } \log_{a} \left(xe^{x}\right)

Answer

exxn−1[xlog⁡ax+1log⁡ea+xlog⁡ax] e^{x} x^{n-1} \left[x \log_{a} x + \frac{1}{\log_{e}a} + x \log_{a}x \right]

Explanation

Solution

ddx[xn(log⁡ax)ex]\frac{d}{dx} \left[x^{n} \left(\log_{a} x\right)e^{x}\right] =xnlog⁡ax.ex+xnex.1xlog⁡ae+nxn−1log⁡ax.ex=x^{n} \log_{a} x. e^{x} +x^{n}e^{x}. \frac{1}{x} \log_{a}e + nx^{n-1} \log_{a}x . e^{x} =ex[xnlog⁡ax+xn−1log⁡ae+nxn−1log⁡ax]= e^{x} \left[x^{n} \log_{a} x+x^{n-1} \log_{a} e + nx^{n-1} \log_{a}x\right] =xn−1ex[xlog⁡ax+1log⁡ea+nlog⁡ax]= x^{ n-1} e^{x} \left[x \log_{a} x + \frac{1}{\log_{e}a} + n \log_{a}x\right]