Differentiation

The standard derivatives Paper 3

Paper 1 gave you $\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=n{x}^{n-1}$ and the chain rule. Paper 3 adds six functions. Learn them, although they are also in the formula list (MF19):

• $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^x)={\mathrm{e}}^x$
• $\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}$, for $x>0$
• $\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x$ and $\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x$
• $\frac{\mathrm{d}}{\mathrm{d}x}(\tan x)={\sec }^{2}x$
• $\frac{\mathrm{d}}{\mathrm{d}x}({\tan }^{-1}x)=\frac{1}{1+x^2}$

Why they are true. $\mathrm{e}$ is the one base for which the gradient of $y=a^x$ at every point equals the height of the curve, so ${\mathrm{e}}^x$ is its own derivative. $\ln x$ is the inverse of ${\mathrm{e}}^x$: if $y=\ln x$ then $x={\mathrm{e}}^y$, so $\frac{\mathrm{d}x}{\mathrm{d}y}={\mathrm{e}}^y=x$, and turning it upside down gives $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{x}$. The same idea gives ${\tan }^{-1}x$: if $y={\tan }^{-1}x$ then $x=\tan y$, so $\frac{\mathrm{d}x}{\mathrm{d}y}={\sec }^{2}y=1+{\tan }^{2}y=1+x^2$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1+x^2}$.

Radians. The trigonometric results are true only when $x$ is in radians. In degrees an extra factor $\frac{\pi }{180}$ appears in every derivative. In calculus, always work in radians, and set your calculator to radians before you evaluate anything.

The derivatives of ${\sin }^{-1}x$ and ${\cos }^{-1}x$ are not required.

Composite functions Paper 3

Combine the standard results with the chain rule, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$. Three patterns come up so often that you should know them on sight:

• $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{\mathrm{f}(x)})=\mathrm{f}\prime (x){\mathrm{e}}^{\mathrm{f}(x)}$, so $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{5x})=5{\mathrm{e}}^{5x}$;
• $\frac{\mathrm{d}}{\mathrm{d}x}\ln \mathrm{f}(x)=\frac{\mathrm{f}\prime (x)}{\mathrm{f}(x)}$, so $\frac{\mathrm{d}}{\mathrm{d}x}\ln (x^2+3)=\frac{2x}{x^2+3}$;
• $\frac{\mathrm{d}}{\mathrm{d}x}\sin (ax+b)=a\cos (ax+b)$, and the same factor $a$ appears for $\cos (ax+b)$ and $\tan (ax+b)$.

Use the laws of logarithms before differentiating where you can. $\ln (\frac{x^3}{2x+1})=3\ln x-\ln (2x+1)$, which differentiates to $\frac{3}{x}-\frac{2}{2x+1}$ with no quotient rule at all. Similarly $\ln (3x)=\ln 3+\ln x$, so its derivative is simply $\frac{1}{x}$.

Powers of trigonometric functions are composites too: ${\sin }^{2}x$ means $(\sin x{)}^2$, so its derivative is $2\sin x\cos x$, by the chain rule with $u=\sin x$.

On this site, type functions with brackets, such as sin(2x), e^(3x) and ln(x^2 + 3). Type ${\sin }^{2}x$ as sin(x)^2, and ${\sec }^{2}x$ as 1/cos(x)^2.

Products and quotients Paper 3

For a product $y=uv$, where $u$ and $v$ are both functions of $x$:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}$$
Why. When $x$ increases by a small amount, $u$ increases by $\delta u$ and $v$ by $\delta v$. The new product is $(u+\delta u)(v+\delta v)=uv+u\delta v+v\delta u+\delta u\delta v$, so $\delta y=u\delta v+v\delta u+\delta u\delta v$. Divide by $\delta x$ and let $\delta x\to 0$: the last term disappears, because it is a product of two small quantities, and the rule is left. The derivative of a product is not the product of the derivatives.

For a quotient $y=\frac{u}{v}$:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$$
The order in the numerator matters: start with $v$, the bottom, or every sign is reversed. The quotient rule is just the product rule applied to $u{v}^{-1}$. Both rules are in MF19.

Write down $u$, $v$, $u\prime$ and $v\prime$ first, then substitute. Factorise the answer, especially when you will set it equal to zero: ${\mathrm{e}}^{2x}(2x^2+2x)$ is far easier to solve than its expanded form, and an exponential factor is never zero.

The rules also build the rest of the toolkit. Since $\tan x=\frac{\sin x}{\cos x}$, the quotient rule gives $\frac{\cos x\cos x-\sin x(-\sin x)}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}={\sec }^{2}x$. In the same way $\frac{\mathrm{d}}{\mathrm{d}x}(\sec x)=\sec x\tan x$, as MF19 states.

Parametric differentiation Paper 3

A curve can be given by parametric equations $x=\mathrm{f}(t)$, $y=\mathrm{g}(t)$: each value of the parameter $t$ gives one point $(x,y)$ of the curve. The chain rule gives its gradient:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}÷\frac{\mathrm{d}x}{\mathrm{d}t}$$
This is in MF19. The gradient comes out in terms of $t$, which is fine: to find the gradient at a point, first find the value of $t$ there.

• Stationary points are where $\frac{\mathrm{d}y}{\mathrm{d}t}=0$ (with $\frac{\mathrm{d}x}{\mathrm{d}t}\ne 0$).
• Where $\frac{\mathrm{d}x}{\mathrm{d}t}=0$ but $\frac{\mathrm{d}y}{\mathrm{d}t}\ne 0$, the tangent is parallel to the $y$-axis.
• Read any restriction on $t$, such as $t>0$. A value of $t$ outside it gives no point on the curve, so discard it.

Finish by giving points as coordinates $(x,y)$, not as values of $t$.

Implicit differentiation Paper 3

Some curves are given by an equation in $x$ and $y$ that is awkward or impossible to rearrange into $y=\dots$, such as $x^2+y^3=4xy$. The curve is then defined implicitly. Differentiate every term with respect to $x$, treating $y$ as a function of $x$:

• a term in $x$ alone differentiates as usual;
• a function of $y$ needs the chain rule: $\frac{\mathrm{d}}{\mathrm{d}x}(y^3)=3y^2\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{2y})=2{\mathrm{e}}^{2y}\frac{\mathrm{d}y}{\mathrm{d}x}$;
• a product of $x$ and $y$ needs the product rule: $\frac{\mathrm{d}}{\mathrm{d}x}(xy)=x\frac{\mathrm{d}y}{\mathrm{d}x}+y$;
• a constant differentiates to 0, on both sides of the equation.

Then collect the terms containing $\frac{\mathrm{d}y}{\mathrm{d}x}$ on one side, factorise, and divide. The gradient is in terms of both $x$ and $y$, so you need both coordinates of a point to evaluate it.

For the points where the tangent is parallel to the $x$-axis, set the numerator of $\frac{\mathrm{d}y}{\mathrm{d}x}$ equal to zero and solve it together with the equation of the curve. On its own, the numerator equal to zero describes a line, not a point. The tangent is parallel to the $y$-axis where the denominator is zero.

Tangents and normals Paper 3

Whichever way the curve is given, the method for a tangent or normal at a point $(x_1,y_1)$ is the one from Paper 1:

1. find the gradient $m$ of the curve at the point, from $\frac{\mathrm{d}y}{\mathrm{d}x}$;
2. the tangent is $y-y_1=m(x-x_1)$, and the normal is $y-y_1=-\frac{1}{m}(x-x_1)$;
3. tidy into the form the question asks for, such as $ax+by+c=0$ with integers.

Keep exact values all the way, such as $\ln 2$ or $\frac{\pi }{4}$ in a coordinate. Rounding a coordinate before writing the line loses the final mark. Exam questions then often use the line: where it meets an axis, or the area of a triangle formed by a tangent, a normal and an axis.