If y = 2^{x }+ x log x, then find dy/dx:

Correct Answer is : 2xlog2+logx+12^{x} log 2 + log x + 12xlog2+logx+1

Correct Answer is : 2xlog2+logx+12^{x} log 2 + log x + 12xlog2+logx+1

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AP DSC School Assistant Math Model Test 1

Section: Subject Specific

Question : 59 of 160

Marks: +1, -0

\frac{dy}{dx}

Solution:

Concept:

Differentiate each term using the standard formulas: for

a^x

\frac{d}{dx}(a^x) = a^x \ln a

; for

x \log x

, use the product rule

\frac{d}{dx}(u v) = u' v + u v'

Explanation:

Given

y = 2^{x} + x \log x

Term 1:

\frac{d}{dx}(2^{x}) = 2^{x} \log 2

Term 2: Apply product rule on

x \log x

\frac{d}{dx}(x \log x) = (1) \cdot \log x + x \cdot \frac{1}{x} = \log x + 1

Combine:

\frac{dy}{dx} = 2^{x} \log 2 + \log x + 1

Answer:

Option B:

2^{x} \log 2 + \log x + 1

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