Differentiation of Polynomial Terms

From a curve to its gradient function

Differentiation finds an expression for the gradient of a curve. If the curve is written as $y=\dots$, its derivative is written as $\frac{dy}{dx}$. This notation means the rate of change of $y$ with respect to $x$.

The original function gives the value of $y$; the derivative gives the gradient. A curve's gradient can change as $x$ changes, so the derivative is often an expression involving $x$, rather than a single number.

Differentiating a power of x

For a term $y=kx^n$, $k$ is a constant coefficient and is the power, or index. Here, can be any integer: positive, zero or negative.

The power rule is:

$$\frac{dy}{dx}=knx^{n-1}$$

There are two changes: multiply the coefficient by the original power, then subtract 1 from the power. For example, if $y=4x^3$,

$$\frac{dy}{dx}=4\times3x^{3-1}=12x^2.$$

A negative coefficient follows exactly the same rule. For $y=-2x^4$, the new coefficient is $-2\times4=-8$, so $\frac{dy}{dx}=-8{}^{}$.

Linear terms and constants

A linear term such as $6x$ has power 1. Differentiating gives $6\times1x^0=6$. This matches the fact that the straight line $y=6x$ has a constant gradient of 6.

A constant differentiates to zero. For example, $y=5$ is a horizontal line: its value does not change as $x$ changes, so its gradient is 0.

Thus the two useful special cases are:

$$y=kx\quad\Rightarrow\quad\frac{dy}{dx}=k$$ $$y=k\quad\Rightarrow\quad\frac{dy}{dx}=0.$$

Differentiating sums and differences

When terms are added or subtracted, differentiate each term separately and combine the results. This is called term-by-term differentiation.

For example, take $y=5x^3+2x+8$. The cubic term becomes $15x^2$, the linear term becomes 2, and the constant becomes 0. Therefore,

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

Treat a subtraction as part of the term's coefficient. For $y=3x^2-7x-6$, the three are , and 0, giving .

Expanding brackets before differentiating

The term-by-term method needs a sum or difference of powers of $x$. If an expression is written as a product of brackets, expand it and collect like terms first.

For $y=(3x+2)(x-3)$, multiply every term in the first bracket by every term in the second:

$$y=3x^2-9x+2x-6=3x^2-7x-6.$$

Now differentiate the simplified expression:

$$\frac{dy}{dx}=6x-7.$$

A squared bracket is also a product. For instance, $(2x+3)^2=(2x+3)(2x+3)=4x^2+12x+9$, so its is .

Rewriting fractions as negative powers

A variable in a denominator can often be rewritten using a negative index. For example,

$$y=\frac{5}{x^3}=5x^{-3},\qquad x\ne0.$$

The power rule still applies. Multiply 5 by , and change the power from to :

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

Both forms represent the same derivative. The restriction remains because the original fraction is undefined at zero.

For a fraction with several numerator terms, divide each term by the denominator before differentiating. For example,

$$y=\frac{8x^6-x^3}{2x^4} =4x^{6-4}-\frac12x^{3-4} =4x^2-\frac12x^{-1},\qquad x\ne0.$$

Differentiating term by term gives

$$\frac{dy}{dx}=8x+\frac12x^{-2}.$$

The second derivative term is positive because its coefficient and its original power are both negative. Whether an expression begins with brackets or fractions, the strategy is the same: rewrite it as separate powers of $x$, then apply the power rule to each term.

Power rule

$$y=kx^n\quad\Rightarrow\quad\frac{dy}{dx}=knx^{n-1}$$