Question

$\begin{aligned} &f(x)=\sin x+\ln x \\ &f\left(x^{2}\right)=\sin x^{2}+\ln x^{2} \\ &2 x f^{\prime}\left(x^{2}\right)=2 x \cos x^{2}+\frac{2}{x} \\ &f^{\prime}\left(x^{2}\right)=\cos x^{2}+\frac{1}{x^{2}} \end{aligned}$

Short answer

Question: Find the function \(f\left(x^{2}\right)\) and its derivative, \(f'\left(x^{2}\right)\), if \(f(x) = \sin{x} + \ln{x}\). Answer: The function \(f(x^2)\) is given by \(\sin{(x^2)} + \ln{(x^2)}\), and its derivative, \(f'(x^2)\), is given by \(2x\left(\cos{(x^2)} + \frac{1}{x^2}\right)\).

Step-by-step solution

Find f(x²)

Substitute \(x^2\) into the function \(f(x)\): $$f(x^2) = \sin{(x^2)} + \ln{(x^2)}$$

Calculate the derivative of f(x²)

In order to compute the derivative of \(f(x^2)\), we must first remember the chain rule of differentiation: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\), where \(y = f(u)\) and \(u = (x^2)\). Differentiate \(f(u) = \sin{u} + \ln{u}\) with respect to \(u\): $$f'(u) = \frac{d(\sin{u})}{du} + \frac{d(\ln{u})}{du} = \cos{u} + \frac{1}{u}$$ Differentiate \(u = x^2\) with respect to \(x\): $$\frac{du}{dx} = \frac{d(x^2)}{dx} = 2x$$ Now by applying the chain rule, find the derivative of \(f(x^2)\) with respect to \(x\): $$f'(x^2) = 2x \left(\cos{(x^2)} + \frac{1} {x^2}\right)$$

Rewrite the answer in the desired form

Finally, we rewrite the answer to match the given expression: $$f'(x^2) = \cos{x^2} + \frac{1}{x^2}$$