Differentiating standard functions

Standard derivatives are the vocabulary of differentiation. Once these results are secure, more complicated functions can be differentiated by combining them with the product, quotient and chain rules.

What you should know first

You should be able to:

• use negative and fractional indices;
• recognise graphs of exponential, logarithmic and trigonometric functions;
• interpret a derivative as an instantaneous rate of change.

Review differentiation and gradients if the meaning of $dy/dx$ or the power rule is unfamiliar. The results below can be justified from first principles, but this lesson focuses on using them accurately.

The essential results

For suitable constants $a$, $c$ and $n$:

$$\begin{aligned} \frac{d}{dx}(c)&=0,\\ \frac{d}{dx}(x^n)&=nx^{n-1},\\ \frac{d}{dx}(e^x)&=e^x,\\ \frac{d}{dx}(a^x)&=a^x\ln a, && a>0,\\ \frac{d}{dx}(\ln x)&=\frac1x, && x>0,\\ \frac{d}{dx}(\sin x)&=\cos x,\\ \frac{d}{dx}(\cos x)&=-\sin x,\\ \frac{d}{dx}(\tan x)&=\sec^2x. \end{aligned}$$

The trigonometric results require $x$ to be measured in radians. This is not a convention that can be ignored: using degrees introduces an extra constant factor.

Two linearity rules let us differentiate expressions term by term:

$$\frac{d}{dx}\bigl(cf(x)\bigr)=cf'(x)$$

and

$$\frac{d}{dx}\bigl(f(x)\pm g(x)\bigr)=f'(x)\pm g'(x).$$

Constants multiply derivatives. Sums and differences remain sums and differences.

Powers, including rational powers

The power rule is

$$\boxed{\frac{d}{dx}(x^n)=nx^{n-1}}.$$

Multiply by the existing power, then subtract one from that power.

Positive integer powers

If

$y=7x^5-3x^2+8x-11,$

then

$$\frac{dy}{dx}=35x^4-6x+8.$$

The derivative of $8x$ is $8$, and the derivative of the constant $-11$ is $0$.

Negative powers

Write reciprocals with indices before differentiating:

$$y=\frac{5}{x^3}-\frac{2}{x}=5x^{-3}-2x^{-1}.$$

Therefore

$$\frac{dy}{dx}=-15x^{-4}+2x^{-2} =-\frac{15}{x^4}+\frac{2}{x^2}.$$

The power decreases by one, so $-3$ becomes $-4$, not $-2$.

Fractional powers

Rewrite roots as powers:

$$\sqrt{x}=x^{1/2},\qquad \sqrt[3]{x^2}=x^{2/3}.$$

For example,

$$y=4\sqrt{x}+3\sqrt[3]{x^2} =4x^{1/2}+3x^{2/3}.$$

Hence

$$\frac{dy}{dx} =2x^{-1/2}+2x^{-1/3} =\frac{2}{\sqrt{x}}+\frac{2}{\sqrt[3]{x}}.$$

The formula works wherever the original function is differentiable as a real function. For example, $\sqrt{x}$ is real only for $x\geq0$, and its derivative $1/(2\sqrt{x})$ is undefined at $x=0$. A correct algebraic rule does not remove domain restrictions.

A mixed example

Differentiate

$$f(x)=\frac{3x^4+2\sqrt{x}-7}{x^2}.$$

Simplify before differentiating:

$$f(x)=3x^2+2x^{-3/2}-7x^{-2}.$$

Now apply the power rule term by term:

$$f'(x)=6x-3x^{-5/2}+14x^{-3}.$$

This is usually quicker and safer than applying the quotient rule to the original expression.

Exponential functions

Why $e^x$ is special

Every exponential function has a rate of change proportional to its current value. The number $e$ is defined so that the proportionality constant is exactly $1$:

$$\boxed{\frac{d}{dx}(e^x)=e^x}.$$

The function and its derivative therefore have the same value at every $x$. At $x=0$, both are $1$.

For a general positive base $a$,

$$a^x=e^{x\ln a}.$$

Differentiating gives

$$\boxed{\frac{d}{dx}(a^x)=a^x\ln a}.$$

The factor $\ln a$ records how the choice of base changes the growth rate. If $a>1$, then $\ln a>0$ and $a^x$ increases. If $0<a<1$, then $\ln a<0$ and $a^x$ decreases.

Example: a sum of exponentials

If

$y=6e^x-4\cdot 3^x+9,$

then

$$\frac{dy}{dx}=6e^x-4\cdot3^x\ln3.$$

There is no $+9$ in the derivative because constants have zero rate of change.

Constant outside or constant inside?

Compare

$$\frac{d}{dx}(5e^x)=5e^x$$

with

$$\frac{d}{dx}(e^{5x})=5e^{5x}.$$

The first uses the constant multiple rule. The second uses the chain rule because $5x$ is inside the exponential. They happen to produce a factor of $5$ for different reasons. For a systematic treatment of functions inside functions, see the product, quotient and chain rules.

Logarithmic functions

Since $y=\ln x$ is the inverse of $y=e^x$, its gradient is the reciprocal of the exponential gradient at the corresponding point. This gives

$$\boxed{\frac{d}{dx}(\ln x)=\frac1x},\qquad x>0.$$

For example, if

$y=4\ln x+3x^2-7,$

then

$$\frac{dy}{dx}=\frac4x+6x.$$

For logarithms to another base,

$$\log_a x=\frac{\ln x}{\ln a},$$

so

$$\boxed{\frac{d}{dx}(\log_a x)=\frac{1}{x\ln a}}.$$

Simplify with logarithm laws first

Consider

$$y=\ln(x^4),\qquad x>0.$$

Using $\ln(x^4)=4\ln x$ gives

$$\frac{dy}{dx}=\frac4x.$$

The domain matters. The expression $\ln(x^4)$ is actually defined for every $x\ne0$, and the derivative is still $4/x$ on both parts of that domain. However, writing $\ln(x^4)=4\ln x$ is valid only for $x>0$. A more careful identity for $x\ne0$ is

$$\ln(x^4)=4\ln|x|.$$

This distinction is useful at A* level and prevents logarithm laws from being applied outside their valid domains.

Do not confuse $\ln x$ with $1/x$

The derivative is

$$\frac{d}{dx}(\ln x)=\frac1x,$$

but the functions are not equal. In particular,

$$\frac{d}{dx}\left(\frac1x\right)=\frac{d}{dx}(x^{-1})=-x^{-2}=-\frac1{x^2}.$$

Trigonometric functions

When angles are in radians,

$$\boxed{\frac{d}{dx}(\sin x)=\cos x}$$

and

$$\boxed{\frac{d}{dx}(\cos x)=-\sin x}.$$

The negative sign in the derivative of cosine is essential. One way to remember the pattern is

$$\sin x\longrightarrow\cos x\longrightarrow-\sin x\longrightarrow-\cos x\longrightarrow\sin x.$$

Also,

$$\boxed{\frac{d}{dx}(\tan x)=\sec^2x}.$$

This follows by writing $\tan x=\sin x/\cos x$ and applying the quotient rule:

$$\frac{d}{dx}(\tan x) =\frac{\cos^2x+\sin^2x}{\cos^2x} =\frac1{\cos^2x} =\sec^2x.$$

Example: a trigonometric polynomial

Differentiate

$y=5\sin x-2\cos x+3\tan x.$

Term by term,

$$\frac{dy}{dx}=5\cos x+2\sin x+3\sec^2x.$$

Notice that differentiating $-2\cos x$ gives $+2\sin x$ because two negative signs combine.

Example: find a tangent gradient

For

$f(x)=3\sin x+4\cos x,$

we have

$f'(x)=3\cos x-4\sin x.$

At $x=\pi/6$,

$$f'\left(\frac\pi6\right) =3\left(\frac{\sqrt3}{2}\right)-4\left(\frac12\right) =\frac{3\sqrt3}{2}-2.$$

Keep exact values unless a decimal approximation is requested.

Why radians matter

Suppose an angle is measured in degrees and define

$$g(x)=\sin\left(\frac{\pi x}{180}\right).$$

Then $g(x)$ represents the sine of $x$ degrees, but

$$g'(x)=\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right).$$

The familiar derivative $\cos x$ appears without an extra factor only when $x$ is in radians.

Combining the standard derivatives

Differentiate

$$y=2x^{5/2}-\frac{3}{x^2}+4e^x-5\ln x+7\sin x-6\cos x.$$

First write the reciprocal as a power:

$$y=2x^{5/2}-3x^{-2}+4e^x-5\ln x+7\sin x-6\cos x.$$

Then differentiate each term:

$$\boxed{ \frac{dy}{dx} =5x^{3/2}+6x^{-3}+4e^x-\frac5x+7\cos x+6\sin x }.$$

Each term requires only one decision: identify its function family, then apply the corresponding standard derivative.

A reliable method

When differentiating a sum of standard functions:

1. Simplify algebraically, especially roots and reciprocals.
2. Separate the expression into terms.
3. Keep each constant coefficient.
4. Apply the correct standard derivative to each term.
5. Check signs, powers and domains.
6. Decide whether any term is composite. If it has a non-trivial inner function, use the chain rule.

Common misconceptions

• Reducing a power in the wrong direction: $d(x^{-3})/dx=-3x^{-4}$, not $-3x^{-2}$.
• Forgetting a coefficient: $d(7\sin x)/dx=7\cos x$.
• Losing the cosine sign: $d(\cos x)/dx=-\sin x$.
• Treating every exponential like $e^x$: $d(2^x)/dx=2^x\ln2$.
• Confusing a function with its derivative: $d(\ln x)/dx=1/x$ does not mean $\ln x=1/x$.
• Using degrees: the standard trigonometric derivatives assume radians.
• Ignoring an inner function: $d(\sin 3x)/dx$ is not simply $\cos3x$. The chain rule supplies a factor of $3$.
• Using the product rule for a constant multiple: $d(5f(x))/dx=5f'(x)$ directly.
• Ignoring domains: a derivative formula is meaningful only where the original function is defined and differentiable.

Check your understanding

Try each question before opening the answer.

1. Powers

Differentiate

$$f(x)=8x^{3/2}-5x^{-1/2}+4.$$

Answer

$$f'(x)=12x^{1/2}+\frac52x^{-3/2}.$$

The constant differentiates to zero. For the second term,

$$-5\left(-\frac12\right)x^{-3/2}=\frac52x^{-3/2}.$$

2. Exponentials and logarithms

Differentiate

$$g(x)=3e^x-2\cdot5^x+7\ln x.$$

Answer

$$g'(x)=3e^x-2\cdot5^x\ln5+\frac7x.$$

3. Trigonometric functions

Differentiate

$$h(x)=4\sin x+9\cos x-2\tan x.$$

Answer

$$h'(x)=4\cos x-9\sin x-2\sec^2x.$$

4. Exact gradient

Given

$$p(x)=2x^3+4\sin x-e^x,$$

find $p'(0)$.

Answer

$$p'(x)=6x^2+4\cos x-e^x,$$

so

$$p'(0)=0+4-1=3.$$

5. Diagnose the error

A student writes

$$\frac{d}{dx}\left(3x^{2/3}-4\cos x+2^x\right) =2x^{-1/3}-4\sin x+2^x.$$

Find every error.

Answer

The power term is correct. The cosine term should become positive because

$$\frac{d}{dx}(-4\cos x)=4\sin x.$$

The exponential needs a factor of $\ln2$. Therefore

$$\frac{d}{dx}\left(3x^{2/3}-4\cos x+2^x\right) =2x^{-1/3}+4\sin x+2^x\ln2.$$

6. A composite function check

Which standard rule is not enough on its own to differentiate $\ln(3x+1)$, and why?

Answer

The derivative of $\ln x$ is not enough on its own because $3x+1$ is an inner function. The chain rule is required:

$$\frac{d}{dx}\ln(3x+1)=\frac{3}{3x+1}.$$

What to learn next

Use these standard derivatives with the product, quotient and chain rules to differentiate products, fractions and composite functions. You can then apply them to tangents and normals and stationary points and curve sketching.