10^{-x tan x} left[ frac{d}{dx} (10^{x tan x} | vedclass

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(B) Let $y = 10^{x 	an x}$.Using the chain rule for differentiation,$\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$.Here,$a = 10$ and $u = x 	an

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$\ln 10 (	an x + x \sec^2 x)$

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(B) Let $y = 10^{x 	an x}$.Using the chain rule for differentiation,$\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$.Here,$a = 10$ and $u = x 	an x$.First,find the derivative of $u = x 	an x$ using the product rule:$\frac{d}{dx}(x 	an x) = \frac{d}{dx}(x) \cdot 	an x + x \cdot \frac{d}{dx}(	an x) = 1 \cdot 	an x + x \cdot \sec^2 x = 	an x + x \sec^2 x$.Now,differentiate $10^{x 	an x}$:$\frac{d}{dx}(10^{x 	an x}) = 10^{x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x)$.Substitute this back into the original expression:$10^{-x 	an x} \left[ 10^{x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x) ight]$$= 10^{-x 	an x + x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x)$$= 10^0 \cdot \ln 10 \cdot (	an x + x \sec^2 x)$$= \ln 10 (	an x + x \sec^2 x)$.

$10^{-x \tan x} \left[ \frac{d}{dx} (10^{x \tan x}) \right]$ is equal to

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