If y = x cdot 7^x, then the value of frac{dx} | vedclass

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(C) Given $y = x \cdot 7^x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot 7^x + x \cdot

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$\frac{1}{7(\log 7 + 1)}$

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(C) Given $y = x \cdot 7^x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot 7^x + x \cdot \frac{d}{dx}(7^x)$$\frac{dy}{dx} = 1 \cdot 7^x + x \cdot (7^x \ln 7)$$\frac{dy}{dx} = 7^x(1 + x \ln 7)$At $x = 1$:$\frac{dy}{dx} = 7^1(1 + 1 \cdot \ln 7) = 7(1 + \ln 7)$Since $\frac{dx}{dy} = \frac{1}{dy/dx}$, we have:$\frac{dx}{dy} = \frac{1}{7(1 + \ln 7)}$Using $\ln 7 = \log_e 7$, the result is $\frac{1}{7(\log 7 + 1)}$.

If $y = x \cdot 7^x$, then the value of $\frac{dx}{dy}$ when $x = 1$ is

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