Question

Find the derivative of logxcos x​

Answer 1

Step-by-step explanation:

Hey mate ,

This is your answer ,

You can find this derivative by applying the Chain Rule, with cosx as the inner function, and lnx as the outer function. Process: To apply the chain rule, we first find the derivative of the outer function, lnu , with u=cosx . Remember that the derivative of lnu=1u=1cosx .

Hope it helps

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Have a great time ahead

Answer 2

Answer:

$\sf{\displaystyle\frac{\cos \left(x\right)-x\sin \left(x\right)\ln \left(x\right)}{\ln \left(10\right)x}}$

Step-by-step explanation:

$\sf{\displaystyle\frac{d}{dx}\left(\log _{10}\left(x\right)\cos \left(x\right)\right)}$

$\sf{\displaystyle=\frac{d}{dx}\left(\log _{10}\left(x\right)\right)\cos \left(x\right)+\frac{d}{dx}\left(\cos \left(x\right)\right)\log _{10}\left(x\right)}$

$\sf{\displaystyle=\frac{1}{x\ln \left(10\right)}\cos \left(x\right)+\left(-\sin \left(x\right)\right)\log _{10}\left(x\right)}$

$\sf{\displaystyle=\frac{1}{x\ln \left(10\right)}\cos \left(x\right)-\sin \left(x\right)\log _{10}\left(x\right)}$

$\sf{\displaystyle=\frac{\cos \left(x\right)}{\ln \left(10\right)x}-\sin \left(x\right)\log _{10}\left(x\right)}$

$\sf{\displaystyle=\frac{\cos \left(x\right)}{x\ln \left(10\right)}-\frac{\sin \left(x\right)\log _{10}\left(x\right)x\ln \left(10\right)}{x\ln \left(10\right)}}$

$\sf{\displaystyle=\frac{\cos \left(x\right)-\sin \left(x\right)\log _{10}\left(x\right)x\ln \left(10\right)}{x\ln \left(10\right)}}$

$\sf{\displaystyle=\frac{\cos \left(x\right)-\frac{\ln \left(x\right)}{\ln \left(e\right)}x\sin \left(x\right)}{\ln \left(10\right)x}}$

$\sf{\displaystyle=\frac{\cos \left(x\right)-x\sin \left(x\right)\ln \left(x\right)}{\ln \left(10\right)x}}$