Continuity and Differentiability - Exponential and Logarithmic Functions

Differentiate $y = e^{\sin(x^2)}$ with respect to $x$.

Given $y = e^{\sin(x^2)}$. Applying the Chain Rule: $\frac{dy}{dx} = \frac{d}{dx}(e^{\sin(x^2)}) = e^{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2))$ $\frac{dy}{dx} = e^{\sin(x^2)} \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2)$ $\frac{dy}{dx} = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x = 2x \cos(x^2) e^{\sin(x^2)}$

We use the derivative rule for $e^u$ which is $e^u \cdot \frac{du}{dx}$, where $u = \sin(x^2)$.

Find $\frac{dy}{dx}$ if $y = \log(\log x), x > 1$.

Let $u = \log x$. Then $y = \log u$. Using the Chain Rule: $\frac{dy}{dx} = \frac{d}{du}(\log u) \cdot \frac{du}{dx}$ $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

The derivative of the outer logarithm is taken first, followed by the derivative of the inner function $\log x$.

Differentiate $y = x^{\sin x}$ with respect to $x$.

Taking natural log on both sides: $\log y = \log(x^{\sin x}) = \sin x \log x$ Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x \log x)$ Using the Product Rule: $\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\sin x)$ $\frac{1}{y} \frac{dy}{dx} = \frac{\sin x}{x} + \log x \cos x$ $\frac{dy}{dx} = y \left( \frac{\sin x}{x} + \cos x \log x \right) = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$

Logarithmic differentiation is necessary here because the variable $x$ appears in both the base and the exponent.

Let $u = \sqrt{x}$. Then $y = e^u$. Using the Chain Rule: $\frac{dy}{dx} = \frac{d}{du}(e^u) \cdot \frac{du}{dx}$ $\frac{dy}{dx} = e^u \cdot \frac{d}{dx}(x^{1/2})$ $\frac{dy}{dx} = e^{\sqrt{x}} \cdot \left(\frac{1}{2}x^{-1/2}\right)$ $\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$

This problem applies the Chain Rule to a composition of the natural exponential function and a square root function. The inner function is $\sqrt{x}$ and the outer function is $e^x$.

First, use the change of base formula to express the log in terms of natural logarithms: $y = \frac{\ln(x^2 + 1)}{\ln 10}$ Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{d}{dx}(\ln(x^2 + 1))$ Using the Chain Rule for the term $\ln(x^2 + 1)$: $\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1)$ $\frac{dy}{dx} = \frac{2x}{(x^2 + 1) \ln 10}$

Standard differentiation rules apply to natural logarithms. For logs with other bases, we convert to base $e$ using $\log_a b = \frac{\ln b}{\ln a}$ before differentiating.