Differentiate \(e^{4x^5}.2x^{log⁡x^2}\) with respect to x.(a) \(e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)(b) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)(c) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5-log⁡2x^2)\)(d) \(x^{log⁡x^2 -1} (10x^4+log⁡2x^2)\)I had been asked this question during an online interview.The origin of the question is Logarithmic Differentiation topic in section Continuity and Differentiability of Mathematics – Class 12

Differentiate \(e^{4x^5}.2x^{log⁡x^2}\) with respect to x.(a) \(e^{4

1.	Differentiate \(e^{4x^5}.2x^{log⁡x^2}\) with respect to x.(a) \(e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)(b) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)(c) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5-log⁡2x^2)\)(d) \(x^{log⁡x^2 -1} (10x^4+log⁡2x^2)\)I had been asked this question during an online interview.The origin of the question is Logarithmic Differentiation topic in section Continuity and Differentiability of Mathematics – Class 12
Answer» RIGHT choice is (b) \(4E^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\) Explanation: Consider y=\(e^{4x^5}+2x^{log⁡x^2}\) Applying log on both sides, we get log⁡y=\(log⁡e^{4x^5} \,+ \,log⁡2x^{log⁡x^2}\) log⁡y=\(4x^5+log⁡x^2 \,. \,log⁡2x\) log⁡y=\(4x^5+2 \,log⁡x \,log⁡2x\) Differentiating with respect to x, we get \(\frac{1}{y} \frac{DY}{dx}\)=\(20x^4+2(\frac{d}{dx} \,(log⁡x) \,log⁡2x+\frac{d}{dx} \,(log⁡2x) \,log⁡x)\) \(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+2\left (\frac{log⁡2x}{x}+\frac{1}{2x}.2.log⁡x\right )\) \(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x+log⁡x)}{x}\) \(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x^2)}{x}\) \(\frac{dy}{dx}\)=\(y(20x^4+\frac{2(log⁡2x^2)}{x})\) \(\frac{dy}{dx}\)=\(e^{4x^5}.2x^{log⁡x^2} (20x^4+\frac{2(log⁡2x^2)}{x})\) \(\frac{dy}{dx}\)=\(4e^{4x^5}.x^{log⁡x^2 -1} (10x^5+log⁡2x^2)\)

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