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$\left(\frac{1}{2}\,\cos^{2}\left(x\right)+\frac{1}{\ln\left(x\right)}\right)'_{x}$

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Dilations of functions

Full solution

Q.

\left(\frac{1}{2}\,\cos^{2}\left(x\right)+\frac{1}{\ln\left(x\right)}\right)'_{x}

Apply Sum Rule: To find the derivative of the function $\left(\frac{1}{2}\cos^{2}(x)+\frac{1}{\ln(x)}\right)$ with respect to $x$ , we will use the sum rule of differentiation, which states that the derivative of a sum is the sum of the derivatives. We will also need to use the chain rule and the power rule for differentiation.
Derivative of First Term: First, let's find the derivative of the first term $\frac{1}{2}\cos^{2}(x)$ . We will use the chain rule, where the outer function is $u^2$ with $u=\cos(x)$ , and the inner function is $\cos(x)$ . The derivative of $u^2$ with respect to $u$ is $2u$ , and the derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$ . Applying the chain rule, we get $u^2$ $0$ .
Derivative of Second Term: Next, we need to find the derivative of the second term $\frac{1}{\ln(x)}$ . We can rewrite this term as $\ln(x)^{-1}$ and then apply the power rule and the chain rule. The derivative of $u^{-1}$ with respect to $u$ is $-u^{-2}$ , and the derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$ . Therefore, the derivative of $\frac{1}{\ln(x)}$ with respect to $x$ is $\ln(x)^{-1}$ $0$ .
Combine Derivatives: Combining the derivatives of both terms, we get the derivative of the entire function: $\newline$ $\left(\frac{1}{2}\cos^{2}(x)+\frac{1}{\ln(x)}\right)' = -\cos(x)\sin(x) - \frac{1}{x\ln(x)^2}$ .

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