If y=(e^{x} · xⁿ · 3^{x}) then \;dy/d x is

Correct Answer is : 1+ nx+log⁡e31+\;n/x+_e 31+xn+loge3

Correct Answer is : 1+ nx+log⁡e31+\;n/x+_e 31+xn+loge3

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Differential Calculus

Question : 9 of 69

Marks: +1, -0

y=\log(e^{x} \cdot x^{n} \cdot 3^{x})

\;\frac{dy}{d x}

Solution:

Concept:

Split the logarithm of a product into a sum of logarithms, then differentiate term by term.

Explanation:

Here

\log

denotes the natural logarithm, i.e.

\log=\log_e

Given

y=\log(e^{x}\cdot x^{n}\cdot 3^{x})

Using

\log(abc)=\log a+\log b+\log c

y=\log(e^{x})+\log(x^{n})+\log(3^{x})

y=x\log_e e+n\log_e x+x\log_e 3

Since

\log_e e=1

, this becomes

y=x+n\log_e x+x\log_e 3

Differentiating with respect to

x

\frac{dy}{dx}=1+n\cdot\frac{1}{x}+\log_e 3

\frac{dy}{dx}=1+\frac{n}{x}+\log_e 3

This matches option B (option B); the other options mis-handle the derivative of

x^{n}

or of the

3^{x}

term.

Answer:

Option B:

1+\frac{n}{x}+\log_e 3

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