4.3 - Differentiating sums of powers of x

4.3 - Differentiating sums of powers of x

Obtain a gradient function directly from a curve's equation. You will differentiate integer powers and their sums, rewrite reciprocal terms as negative powers and simplify products or fractions before applying the same rule.

Differentiating powers and constants

Could moving an entire curve upwards by a million units leave every gradient unchanged? Keep that question in mind: a curve's height and its slope are different information.

You already know

The gradient function dydx\frac{\mathrm{d}y}{\mathrm{d}x} gives the tangent's gradient. Substituting into that function gives a local slope; substituting into yy gives a height.

Differentiation is the process of obtaining the gradient function. For a power of xx, multiply the coefficient by the old power, then reduce that power by one. For example, x2x^2 has gradient function 2x2x, and 3x43x^4 has gradient function 12x312x^3.

The integer power rule

For y=kxny=kx^n, where kk is constant and nn is a non-zero integer,

dydx=knxn−1\frac{\mathrm{d}y}{\mathrm{d}x}=knx^{n-1}

For negative nn, require x≠0x\ne0. A constant y=ky=k has derivative 00. Differentiate a sum or difference term by term.
Not on the formulae sheet: learn it.

The special cases fit familiar gradients: y=5xy=5x has derivative 5x0=55x^0=5, while a constant function is horizontal and has derivative 00. Simplify the first result to the constant 55, which also gives the gradient at x=0x=0; you do not need to evaluate 000^0. A non-zero constant can be written kx0kx^0, but handle its derivative as zero rather than writing an unnecessary negative power.

Worked example 1: a sum with a constant

Question. Given y=4x3−5x+7y=4x^3-5x+7, work out dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

Plan. Differentiate the cubic, linear and constant terms separately.

dydx=4(3)x3−1−5(1)x1−1+0apply the rule to each term=12x2−5simplify x0=1\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=4(3)x^{3-1}-5(1)x^{1-1}+0 &&\text{apply the rule to each term}\\ &=12x^2-5 &&\text{simplify }x^0=1 \end{aligned}

Why this step? A constant contributes no change as xx changes, so the +7+7 contributes zero gradient.

Answer. dydx=12x2−5\frac{\mathrm{d}y}{\mathrm{d}x}=12x^2-5.

Check. The cubic term produces a quadratic term. At x=0x=0, the derivative is −5-5, matching the slope of the linear term there; the cubic's derivative contributes zero there.

Change only the final constant from +7+7 to −9-9. The heights change, but watch what happens to the gradient function.

Worked example 2: shifting a curve vertically

Question. Given y=4x3−5x−9y=4x^3-5x-9, work out dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

Plan. Differentiate each term, including the new constant.

dydx=4(3)x2−5+0constant still gives zero=12x2−5collect\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=4(3)x^2-5+0 &&\text{constant still gives zero}\\ &=12x^2-5 &&\text{collect} \end{aligned}

Why this step? Replacing one constant by another shifts every point vertically by the same amount, without changing any tangent gradient.

Answer. dydx=12x2−5\frac{\mathrm{d}y}{\mathrm{d}x}=12x^2-5, exactly as before.

Check. At x=1x=1, both curves have gradient 77, although their heights are 66 and −10-10, differing by 1616.

This answers the opening question: adding 1 000 0001\,000\,000 changes every height by a million and every gradient by zero. Indeed, when you subtract two heights to form a gradient, the added constants cancel. Differentiation deliberately loses information about a curve's vertical position.

Watch out

Wrong: the derivative of 4x34x^3 is 4x24x^2. Reducing the power is only half the rule. Multiply by the old power too: 4(3)x2=12x24(3)x^2=12x^2.

Differentiate each term in the new sum; leave no constant behind.

Differentiating reciprocal terms

A letter in the denominator becomes a negative power. The same power rule then works, including its instruction to subtract one from the exponent.

You already know

For x≠0x\ne0, kxr=kx−r\frac{k}{x^r}=kx^{-r}. The coefficient stays outside the power: 34x2=34x−2\frac3{4x^2}=\frac34x^{-2}, not 12x−212x^{-2}.

For a negative exponent, subtracting one makes it more negative: −3−1=−4-3-1=-4. Keep coefficient multiplication and exponent subtraction as separate decisions.

Worked example 3: two reciprocal terms

Question. Given y=5x3+2xy=\dfrac5{x^3}+\dfrac2x, work out dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

Plan. Rewrite both terms with negative powers, then differentiate separately.

y=5x−3+2x−1rewrite reciprocalsdydx=5(−3)x−4+2(−1)x−2multiply and lower powers=−15x−4−2x−2simplify coefficients\begin{aligned} y&=5x^{-3}+2x^{-1} &&\text{rewrite reciprocals}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=5(-3)x^{-4}+2(-1)x^{-2} &&\text{multiply and lower powers}\\ &=-15x^{-4}-2x^{-2} &&\text{simplify coefficients} \end{aligned}

Why this step? The original expression is a sum of powers once rewritten, so the same rule applies without treating the denominator as a separate function.

Answer. dydx=−15x4−2x2\frac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{15}{x^4}-\dfrac2{x^2}, for x≠0x\ne0.

Check. At x=1x=1, the derivative is −17-17. Both derivative terms are negative wherever they are defined, because their denominators are positive even powers.

Change only the sign of the second term. Its negative coefficient now multiplies a negative exponent.

Worked example 4: a negative reciprocal term

Question. Given y=5x3−2xy=\dfrac5{x^3}-\dfrac2x, work out dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

Plan. Keep the coefficient −2-2 attached while applying the rule.

y=5x−3−2x−1rewritedydx=5(−3)x−4+(−2)(−1)x−2keep signed coefficients=−15x−4+2x−2negative times negative\begin{aligned} y&=5x^{-3}-2x^{-1} &&\text{rewrite}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=5(-3)x^{-4}+(-2)(-1)x^{-2} &&\text{keep signed coefficients}\\ &=-15x^{-4}+2x^{-2} &&\text{negative times negative} \end{aligned}

Why this step? The exponent is still reduced from −1-1 to −2-2; the positive sign comes from multiplying two negatives, not from changing that exponent rule.

Answer. dydx=−15x4+2x2\frac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{15}{x^4}+\dfrac2{x^2}, for x≠0x\ne0.

Check. At x=1x=1, the derivative is −13-13: the changed term contributes +2+2, rather than −2-2.

Negative powers do not introduce a new calculus rule; they demand extra care with the arithmetic already inside the rule.

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