P1.4 - Differentiation

Learning objectives

P1.4.1Derivative as gradient and rate of changeThe derivative of f(x) as the dy For example, knowledge that is the rate of change of y gradient of the tangent to the graph dx of y = f(x) at a point; the gradient of with respect to x.; Knowledge of the chain rule is not the tangent as a limit; interpretation required. as a rate of change; second order derivatives.; The notation f ′(x) and f ′′(x) may be used.P1.4.2Differentiating powers of xDifferentiation of xn, and related The ability to differentiate expressions such as sums, differences and constant x2 +5x −3 (2x + 5)(x − 1) and is expected. multiples. 3 x.P1.4.3Tangents, normals and gradientsApplications of differentiation to Use of differentiation to find equations of tangents and gradients, tangents and normals. normals at specific points on a curve.

For $y=f(x)$ , the derivative $f'(x)$ or $\dfrac{dy}{dx}$ gives the instantaneous rate of change of $y$ with respect to $x$ . Geometrically, its value at $x=a$ is the gradient of the tangent to the curve there.

$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$

The quotient is the gradient of the secant through inputs $a$ and $a+h$ . As $h$ tends to zero, the second point approaches the first and the secant gradients approach the tangent gradient. This limit is why a derivative describes change at one point rather than only an average over an interval.

For $f(x)=x^2$ , $\dfrac{f(a+h)-f(a)}h=\dfrac{(a+h)^2-a^2}{h}=2a+h$ . Taking $h\to0$ gives $f'(a)=2a$ , so the tangent gradient at $x=3$ is $6$ .

Differentiating again gives $f''(x)=\dfrac{d^2y}{dx^2}$ , the rate at which the first derivative changes. If $y$ is displacement and $x$ is time, for example, $dy/dx$ is velocity and $d^2y/dx^2$ is acceleration; the units change accordingly.

A derivative is a limit of nearby secant gradients, not the gradient between two fixed points. An instantaneous rate is obtained by evaluating the derivative at the required input.

$\frac{d}{dx}(x^n)=nx^{n-1},\qquad \frac{d}{dx}(c)=0$

The power rule applies term by term to sums, differences and constant multiples. Before using it, expand products and rewrite roots or variables in denominators as powers of $x$ ; this exposes every term in the required form $ax^n$ .

Original structure	Preparation
$(2x+5)(x-1)$	expand to $2x^2+3x-5$
$\sqrt{x}$	write as $x^{1/2}$
$1/x^p$	write as $x^{-p}$
a quotient over $\sqrt{x}$	divide each numerator term by $x^{1/2}$

For $y=\dfrac{x^2+5x-3}{3\sqrt{x}}$ , first write $y=\tfrac13x^{3/2}+\tfrac53x^{1/2}-x^{-1/2}$ . Therefore $\dfrac{dy}{dx}=\tfrac12x^{1/2}+\tfrac56x^{-1/2}+\tfrac12x^{-3/2}$ . Each coefficient is multiplied by its old power, then the power is reduced by one.

A constant differentiates to zero: expanding $(2x+5)(x-1)$ gives derivative $4x+3$ . Preserve any original domain restriction; the quotient example still requires $x>0$ even after it is rewritten.

At a point on a curve, differentiation supplies the tangent gradient. The normal is the straight line through the same point that is perpendicular to that tangent.

Step	Action
locate the point	use the given coordinates, or substitute its $x$ -value into the curve
tangent gradient	find $f'(x)$ and evaluate $m_T=f'(a)$
normal gradient	use $m_N=-1/m_T$ when $m_T\ne0$
equation	substitute the point into $y-y_1=m(x-x_1)$

For $y=x^2+2x$ at $x=1$ , the point is $(1,3)$ and $dy/dx=2x+2$ , so $m_T=4$ . The tangent is $y-3=4(x-1)$ . The normal gradient is $-\tfrac14$ , so the normal is $y-3=-\tfrac14(x-1)$ . Substitution of $(1,3)$ checks both equations pass through the correct point.

If a tangent must be parallel to a line of gradient $m$ , solve $f'(x)=m$ for every valid $x$ , then recover the corresponding point on the curve. For a perpendicular tangent, equate $f'(x)$ to the negative reciprocal of the given line's gradient.

Do not use $f'(x)$ as a line equation: it supplies a gradient only after evaluation at the point. When $m_T=0$ , the tangent is horizontal and its normal is the vertical line $x=a$ , so the negative-reciprocal formula is not finite.

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P1.4 - Differentiation

Learning objectives

A derivative is the limiting gradient at one point

Rewrite first, then differentiate each power of x

Turn a derivative value into a tangent or normal

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