Differentiation

The standard derivatives Paper 2

Paper 1 differentiated powers of $x$ and used the chain rule on expressions such as $(3x+2{)}^5$. Paper 2 adds five functions. Learn them; they are also printed in the formula list (MF19).
$$\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^x)={\mathrm{e}}^x\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x\frac{\mathrm{d}}{\mathrm{d}x}(\tan x)={\sec }^{2}x$$
Why they are true:

• ${\mathrm{e}}^x$ is the one exponential function whose gradient at every point equals its height. That property is what defines $\mathrm{e}$, and it is why ${\mathrm{e}}^x$ appears whenever a quantity grows at a rate proportional to its size.
• $\ln x$ is the inverse of ${\mathrm{e}}^x$. If $y=\ln x$ then $x={\mathrm{e}}^y$, so $\frac{\mathrm{d}x}{\mathrm{d}y}={\mathrm{e}}^y=x$. Turning this upside down gives $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{x}$. It holds for $x>0$, where $\ln x$ is defined.
• $\sin x$ and $\cos x$: the gradient of $y=\sin x$ is 1 at $x=0$, zero at $x=\frac{1}{2}\pi$ and $-1$ at $x=\pi$, exactly the values of $\cos x$. These results hold only when $x$ is in radians; in degrees every derivative would carry an extra factor $\frac{\pi }{180}$. In calculus, always work in radians.
• $\tan x$ follows from the quotient rule, below.

Constant multiples and sums work as before: $\frac{\mathrm{d}}{\mathrm{d}x}(5{\mathrm{e}}^x-3\ln x+2\cos x)=5{\mathrm{e}}^x-\frac{3}{x}-2\sin x$.

Composite functions Paper 2

The chain rule, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$, works with the new functions exactly as with powers: differentiate the outer function, leave the inside alone, then multiply by the derivative of the inside. Three patterns come up constantly:
$$\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{\mathrm{f}(x)})=\mathrm{f}\prime (x){\mathrm{e}}^{\mathrm{f}(x)}\frac{\mathrm{d}}{\mathrm{d}x}\ln \mathrm{f}(x)=\frac{\mathrm{f}\prime (x)}{\mathrm{f}(x)}\frac{\mathrm{d}}{\mathrm{d}x}\sin (ax+b)=a\cos (ax+b)$$
So $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{5x})=5{\mathrm{e}}^{5x}$, $\frac{\mathrm{d}}{\mathrm{d}x}\ln (x^2+4)=\frac{2x}{x^2+4}$ and $\frac{\mathrm{d}}{\mathrm{d}x}(\cos 3x)=-3\sin 3x$.

Powers of trigonometric functions are composites too. ${\sin }^{2}x$ means $(\sin x{)}^2$, so its derivative is $2\sin x\times \cos x=2\sin x\cos x$.

Simplify logarithms first. The laws of logarithms from topic 2.2 often remove most of the work. $\ln 3x=\ln 3+\ln x$ has derivative $\frac{1}{x}$, because $\ln 3$ is a constant. And $\ln \frac{(x+1{)}^3}{x}=3\ln (x+1)-\ln x$ differentiates term by term to $\frac{3}{x+1}-\frac{1}{x}$, with no quotient rule needed.

In the answer boxes on this site, put a whole power or angle in brackets: type ${\mathrm{e}}^{2x}$ as e^(2x), $\sin 3x$ as sin(3x) and ${\sin }^{2}x$ as sin(x)^2.

Products Paper 2

For $y=uv$, 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: picture $uv$ as the area of a rectangle with sides $u$ and $v$. When $x$ increases by $\delta x$, $u$ grows by $\delta u$ and $v$ by $\delta v$. The area gains a strip $u\delta v$, a strip $v\delta u$ and a tiny corner $\delta u\delta v$. Divide by $\delta x$ and let $\delta x\to 0$: the corner vanishes and the two strips give the rule.

The derivative of a product is not the product of the derivatives. Write down $u$, $v$, $\frac{\mathrm{d}u}{\mathrm{d}x}$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ first, then assemble. For the syllabus example $x{\mathrm{e}}^{1-x^2}$: $u=x$ and $v={\mathrm{e}}^{1-x^2}$, so the derivative is $x\times (-2x){\mathrm{e}}^{1-x^2}+{\mathrm{e}}^{1-x^2}=(1-2x^2){\mathrm{e}}^{1-x^2}$.

Factorise where you can, because the next part of a question usually sets the derivative equal to zero. Since ${\mathrm{e}}^{\mathrm{f}(x)}>0$ for every $x$, an exponential factor can never be zero, and the stationary points come from the other factor alone: here $1-2x^2=0$, so $x=\pm \frac{1}{\sqrt{2}}$.

Quotients Paper 2

For $y=\frac{u}{v}$:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$$
It comes from the product rule applied to $u{v}^{-1}$, which gives $u\times (-v^{-2}\frac{\mathrm{d}v}{\mathrm{d}x})+v^{-1}\frac{\mathrm{d}u}{\mathrm{d}x}$, written over the common denominator $v^2$. Because of the minus sign the order matters: start with $v$, the bottom.

The syllabus example $y=\frac{2x-4}{3x+2}$ has $u=2x-4$ and $v=3x+2$, so
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(3x+2)(2)-(2x-4)(3)}{(3x+2{)}^2}=\frac{16}{(3x+2{)}^2}$$
The derivative of $\tan x$ comes the same way. With $\tan x=\frac{\sin x}{\cos x}$:
$$\frac{\mathrm{d}}{\mathrm{d}x}(\tan x)=\frac{\cos x\cos x-\sin x(-\sin x)}{{\cos }^{2}x}=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}={\sec }^{2}x$$
Sometimes a quotient can be avoided: $\frac{{\mathrm{e}}^x}{{\mathrm{e}}^{3x}}={\mathrm{e}}^{-2x}$, and $\frac{5}{(2x+1{)}^3}$ is simply a power. Use the quotient rule when the bottom is a genuine function you cannot simplify away.

Parametric equations Paper 2

A curve can be given by two equations, $x=\mathrm{f}(t)$ and $y=\mathrm{g}(t)$, in terms of a parameter $t$. Each value of $t$ gives one point $(x,y)$. To find the gradient you do not need to eliminate $t$. By the chain rule,
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}÷\frac{\mathrm{d}x}{\mathrm{d}t}$$
This is printed in MF19. The gradient comes out in terms of $t$. To use it at a point, find the value of $t$ there first, then substitute that value into the coordinates and into the gradient.

For the syllabus example $x=t-{\mathrm{e}}^{2t}$, $y=t+{\mathrm{e}}^{2t}$: $\frac{\mathrm{d}x}{\mathrm{d}t}=1-2{\mathrm{e}}^{2t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}=1+2{\mathrm{e}}^{2t}$, so $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+2{\mathrm{e}}^{2t}}{1-2{\mathrm{e}}^{2t}}$.

• The tangent is parallel to the $x$-axis where $\frac{\mathrm{d}y}{\mathrm{d}t}=0$ (with $\frac{\mathrm{d}x}{\mathrm{d}t}\ne 0$). Here $1+2{\mathrm{e}}^{2t}$ is always positive, so there are no such points.
• The tangent is parallel to the $y$-axis where $\frac{\mathrm{d}x}{\mathrm{d}t}=0$. Here that needs ${\mathrm{e}}^{2t}=\frac{1}{2}$, so $t=-\frac{1}{2}\ln 2$.

Implicit differentiation Paper 2

An equation such as $x^2+y^2=xy+7$ defines a curve without giving $y$ as a function of $x$: it defines $y$ implicitly. Differentiate every term with respect to $x$, treating $y$ as a function of $x$:

• a term in $x$ alone differentiates as usual;
• a term in $y$ needs the chain rule: $\frac{\mathrm{d}}{\mathrm{d}x}(y^2)=2y\frac{\mathrm{d}y}{\mathrm{d}x}$, $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^y)={\mathrm{e}}^y\frac{\mathrm{d}y}{\mathrm{d}x}$, $\frac{\mathrm{d}}{\mathrm{d}x}(\sin y)=\cos y\frac{\mathrm{d}y}{\mathrm{d}x}$;
• 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 gives 0.

Then collect the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms on one side, factorise and divide. For the syllabus example:
$$2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=x\frac{\mathrm{d}y}{\mathrm{d}x}+y$$
so $(2y-x)\frac{\mathrm{d}y}{\mathrm{d}x}=y-2x$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y-2x}{2y-x}$.

The gradient is in terms of both $x$ and $y$, so at a point you need both coordinates. Check first that the point satisfies the equation of the curve: it catches copying errors.

Tangents and normals. With gradient $m$ at $(x_1,y_1)$, the tangent is $y-y_1=m(x-x_1)$, and the normal, perpendicular to it, has gradient $-\frac{1}{m}$. This applies to every kind of curve in this topic. A tangent is parallel to the $x$-axis where the numerator of $\frac{\mathrm{d}y}{\mathrm{d}x}$ is zero; solve that equation together with the equation of the curve.