Differentiation II

Unit 1 differentiated $y = f(x)$. Here the function may be tangled up with $y$, or defined through a parameter, or exponential — and each case has its own technique.

Exponential and logarithmic derivatives

$$\frac{d}{dx}\left(e^x\right) = e^x$$

The exponential function is its own derivative — the property that defines $e$.

With the chain rule:

$$\frac{d}{dx}\left(e^{kx}\right) = ke^{kx} \qquad \frac{d}{dx}\left(e^{f(x)}\right) = f'(x)e^{f(x)}$$

$$\frac{d}{dx}\left(e^{3x}\right) = 3e^{3x} \qquad \frac{d}{dx}\left(e^{x^2}\right) = 2xe^{x^2}$$

For logarithms:

$$\frac{d}{dx}(\ln x) = \frac1x \qquad \frac{d}{dx}\big(\ln f(x)\big) = \frac{f'(x)}{f(x)}$$

$$\frac{d}{dx}\ln(3x^2+1) = \frac{6x}{3x^2+1}$$

And for a general base:

$$\frac{d}{dx}\left(a^x\right) = a^x\ln a$$

Combining rules. For $y = x\ln x$, the product rule gives

$$\frac{dy}{dx} = (1)(\ln x) + (x)\left(\frac1x\right) = \ln x + 1$$

Implicit differentiation

Some curves cannot be written as $y = f(x)$ — a circle fails the vertical line test. Differentiate every term with respect to $x$, using the chain rule whenever $y$ appears:

$$\frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx} \qquad \frac{d}{dx}\left(y^3\right) = 3y^2\frac{dy}{dx}$$

Differentiate $x^2 + y^2 = 25$ implicitly.

$$2x + 2y\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y}$$

At $(3,4)$ the gradient is $-\frac34$. Check geometrically: the radius to that point has gradient $\frac43$, and a tangent is perpendicular to the radius ✓

A product of $x$ and $y$ needs the product rule.

$$\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$$

So for $x^2 + xy + y^2 = 7$:

$$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx}\big(x + 2y\big) = -(2x + y)$$ $$\frac{dy}{dx} = -\frac{2x+y}{x+2y}$$

Parametric differentiation

When $x$ and $y$ are both given in terms of a parameter $t$:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$

For $x = t^2$ and $y = t^3$:

$$\frac{dx}{dt} = 2t \qquad \frac{dy}{dt} = 3t^2$$ $$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$$

For $x = 2\cos t$, $y = 3\sin t$ (an ellipse):

$$\frac{dy}{dx} = \frac{3\cos t}{-2\sin t} = -\frac32\cot t$$

Inverse trigonometric derivatives

$$\frac{d}{dx}\left(\sin^{-1}x\right) = \frac{1}{\sqrt{1-x^2}}$$ $$\frac{d}{dx}\left(\cos^{-1}x\right) = -\frac{1}{\sqrt{1-x^2}}$$ $$\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1+x^2}$$

Notice that the first two differ only in sign, and that none of the three contains a trigonometric function — which is why they turn up as answers to integration questions involving surds and quadratics.

With the chain rule:

$$\frac{d}{dx}\tan^{-1}(2x) = \frac{2}{1+4x^2}$$

Higher derivatives

Differentiate repeatedly:

$$y = e^{2x} \ \Rightarrow \ \frac{dy}{dx} = 2e^{2x} \ \Rightarrow \ \frac{d^2y}{dx^2} = 4e^{2x}$$

Each differentiation of $e^{kx}$ brings down another factor of $k$, so $\frac{d^ny}{dx^n} = k^ne^{kx}$.

For $y = \ln x$:

$$\frac{dy}{dx} = x^{-1} \qquad \frac{d^2y}{dx^2} = -x^{-2} \qquad \frac{d^3y}{dx^3} = 2x^{-3}$$

Applications

Everything from Unit 1 still applies — the new derivatives simply widen the range of curves.

Stationary points. For $y = xe^{-x}$:

$$\frac{dy}{dx} = e^{-x} - xe^{-x} = e^{-x}(1-x) = 0$$

Since $e^{-x}$ is never zero, $x = 1$, giving $y = e^{-1}$. The second derivative is $e^{-x}(x-2)$, which at $x=1$ is $-e^{-1} < 0$ — a maximum at $\left(1, \frac1e\right)$.

Tangents to implicit curves. Find $\frac{dy}{dx}$ implicitly, substitute the point's coordinates (both of them), and use $y - y_1 = m(x-x_1)$ as usual.