Calculus - The chain rule

Find the derivative of $f(x) = (3x^2 + 5)^7$.

$f'(x) = 7(3x^2 + 5)^6 \cdot \frac{d}{dx}(3x^2 + 5)$ $f'(x) = 7(3x^2 + 5)^6 \cdot (6x)$ $f'(x) = 42x(3x^2 + 5)^6$

We identify the 'outer' function as $u^7$ and the 'inner' function as $u = 3x^2 + 5$. Applying the chain rule, we differentiate the power first, then multiply by the derivative of the polynomial inside.

Differentiate $y = e^{4x^3 - 2x}$ with respect to $x$.

$\frac{dy}{dx} = e^{4x^3 - 2x} \cdot \frac{d}{dx}(4x^3 - 2x)$ $\frac{dy}{dx} = e^{4x^3 - 2x} \cdot (12x^2 - 2)$ $\frac{dy}{dx} = (12x^2 - 2)e^{4x^3 - 2x}$

The derivative of $e^u$ is $e^u \cdot \frac{du}{dx}$. Here, $u = 4x^3 - 2x$. We multiply the original exponential term by the derivative of the exponent.

Find $\frac{dy}{dx}$ for $y = \ln(\sin(x))$.

$\frac{dy}{dx} = \frac{1}{\sin(x)} \cdot \frac{d}{dx}(\sin(x))$ $\frac{dy}{dx} = \frac{1}{\sin(x)} \cdot \cos(x)$ $\frac{dy}{dx} = \cot(x)$

Using the chain rule for logarithms, $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}$. Here $u = \sin(x)$, so we multiply $\frac{1}{\sin(x)}$ by the derivative of $\sin(x)$, which is $\cos(x)$.