Differentiation (Edexcel IGCSE Maths)

Spec point — 3.4C

Differentiation

What is a gradient function?

• Recall that the equation of a curve gives the y-coordinate of a point when you substitute in its x-coordinate

  • For example, $y=x^{2}+3x+5$

    • Substitute $x=2$ in to get $y=2^{2}+3\times2+5=15$
    • The point $( 2 , 15 )$ lies on the curve

• A gradient function gives the gradient of the curve at a point when you substitute in its x-coordinate

  • A gradient function is written as $\frac{dy}{dx}=...$

    • pronounced "dy by dx"
    • This refers to the change in y over change in x

  • For example, the gradient function $\frac{dy}{dx}=2x+3$

    • Substitute $x=2$ in to get $\frac{dy}{dx}=2\times2+3=7$
    • The gradient at the point where $x=2$ is 7

• The gradient function $\frac{dy}{dx}$ can also be called the derivative or derived function

What is differentiation and how does it work?

• Differentiation is an algebraic method that changes the equation of a curve, $y=...$, into a gradient function, $\frac{dy}{dx}=...$

• To differentiate a power of $x$, bring down the power and reduce the power by 1

  • Differentiating $y=x^{5}$ gives $\frac{dy}{dx}=5x^{4}$
  • Differentiating $y=x^{8}$ gives $\frac{dy}{dx}=8x^{7}$
  • Differentiating $y=x^{n}$ gives $\frac{dy}{dx}=nx^{n-1}$

• Any number that is already in front of $x$ is multiplied by the power that was brought down

  • Differentiating $y=2x^{5}$ gives $\frac{dy}{dx}=2\times5x^{4}=10x^{4}$
  • Differentiating $y=\frac{1}{2}x^{8}$ gives $\frac{dy}{dx}=\frac{1}{2}\times8x^{7}=4x^{7}$
  • Differentiating $y=kx^{n}$ gives $\frac{dy}{dx}=knx^{n-1}$

• Be careful with two special cases:

  • Differentiating $y=kx$ gives $\frac{dy}{dx}=k$

    • The $x$ disappears leaving just a number
    • For example, $y=2x$ differentiates to $\frac{dy}{dx}=2$
    • This makes sense, the gradient of the straight line $y=2x$ is 2

  • Differentiating just a number (a constant term), $y=c$, gives $\frac{dy}{dx}=0$

    • Numbers disappear!
    • For example, $y=4$ differentiates to $\frac{dy}{dx}=0$
    • This makes sense, the gradient of the horizontal line $y=4$ is zero

How do I differentiate negative powers of x?

• The same rules apply to negative powers, $y=x^{n}$ becomes $\frac{dy}{dx}=nx^{n-1}$

  • Be careful: subtracting 1 from a negative power creates a larger negative number!

    • E.g. $y=x^{-3}$ becomes $\frac{dy}{dx}=-3x^{-3-1}=-3x^{-4}$

• You may need to use index laws to convert algebraic fractions into negative powers (and vice versa)

  • For example

    • $y=\frac{1}{x}\rightarrow y=x^{-1}$ so $\frac{dy}{dx}=-x^{-2}=-\frac{1}{x^{2}}$
    • $y=\frac{5}{x^{6}}\rightarrow y=5x^{-6}$ so $\frac{ \mathrm{d} y}{ \mathrm{d} x} = 5 \times ( - 6 ) x^{- 7} = - \frac{30}{x^{7}}$
    • $y=\frac{4}{3x^{10}}\rightarrow y=\frac{4}{3}x^{-10}$ so $\frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{4}{3} \times ( - 10 ) x^{- 11} = - \frac{40}{3 x^{11} }$

How do I differentiate sums and differences of terms?

• The equation of a curve may include a number of different terms

  • You can differentiate each term individually

    • Differentiating $y=x^{5}+x^{8}$ gives $\frac{dy}{dx}=5x^{4}+8x^{7}$
    • Differentiating $y=4x^{3}-10x^{6}$ gives $\frac{dy}{dx}=12x^{2}-60x^{5}$

• Remember the two special cases of:

  • $kx$ differentiating to just $k$

  • and constant terms, $c$, differentiating to zero

    • E.g. differentiating $y=8x+1$ gives $\frac{dy}{dx}=8$

• You may need to separate an algebraic fraction

  • E.g. $y=\frac{x^{3}+2x}{x^{5}}$ becomes $y=\frac{x^{3}}{x^{5}}+\frac{2x}{x^{5}}=\frac{1}{x^{2}}+\frac{2}{x^{4}}$

    • Using negative indices gives $y=x^{-2}+2x^{-4}$
    • This can be differentiated: $\frac{ \mathrm{d} y}{ \mathrm{d} x} = - 2 x^{- 3} + 2 ( - 4 ) x^{- 5}$
    • This simplifies to $\frac{dy}{dx}=-\frac{2}{x^{3}}-\frac{8}{x^{5}}$

How do I find the gradient of a curve using the gradient function?

• Find the x-coordinate of the point on the curve you're interested in

• Use differentiation to turn the equation of the curve, $y=...$, into the gradient function, $\frac{dy}{dx}=...$

• Substitute the x-coordinate into the gradient function to find the gradient

  • The y-coordinate is not needed

• If instead you are given a gradient and asked to find the x-coordinate

  • set $\frac{dy}{dx}$ equal to that gradient and solve the equation

Exam Hint Don't forget to write the left-hand sides of $y=...$ and $\frac{dy}{dx}=...$ to avoid mixing up the curve equation with the gradient function!

How is the gradient function related to drawing tangents?

• The gradient of a curve changes as you move along the curve

  • To find the gradient at a particular point you can draw a tangent and find its gradient

  • This is a graphical method that is not accurate

    • It depends on how well you draw the tangent

• Instead, you can use differentiation to find the gradient function, then substitute the x-coordinate of the point into the gradient function to find the gradient

  • This is an algebraic method that is exact

• For example, to find the gradient of the curve $y=x^{3}-2x^{2}+6$ at the point P where $x=2$

  • either try to draw a tangent at $x=2$ and measure its gradient (see below)

  • or differentiate the equation to get $\frac{dy}{dx}=3x^{2}-4x$

    • Substitute in $x=2$ to get $\frac{dy}{dx}=3\times2^{2}-4\times2=4$
    • The gradient is 4

Worked Example A curve has the equation $y=x^{3}-3x^{2}-4x+1$.

(a) Find the gradient of the curve at the point $( 1 , - 5 )$.

Answer:

Find the gradient function using differentiation

$\begin{gathered} \frac{ \mathrm{d} y}{ \mathrm{d} x} & = 3 x^{2} - 3 \times 2 x - 4 + 0 \\ \frac{ \mathrm{d} y}{ \mathrm{d} x} & = 3 x^{2} - 6 x - 4 \end{gathered}$

Substitute $x=1$ into the gradient function

$\begin{gathered} \frac{ \mathrm{d} y}{ \mathrm{d} x} & = 3 \times 1^{2} - 6 \times 1 - 4 \\ = - 7 \end{gathered}$

The gradient when $x=1$ is -7 (The y-coordinate of the point is not needed)

Final answer: The gradient at$(1,-5)$is -7

(b) Find the coordinates of the two points on the curve with a gradient of 5.

Answer:

This is saying that $\frac{dy}{dx}=5$ (This is not the same as substituting in $x=5$)

Form an equation using $\frac{dy}{dx}$ from above

$3x^{2}-6x-4=5$

This is a quadratic equation Bring the terms to one side and solve (for example, by factorisation)

$\begin{gathered} 3 x^{2} - 6 x - 9 & = 0 \\ x^{2} - 2 x - 3 & = 0 \\ ( x - 3 ) ( x + 1 ) & = 0 \\ x = 3 \text{or} x = - 1 \end{gathered}$

These are the $x$-coordinates of the two points on the curve with gradient 5 Substitute $x=3$ into $y=x^{3}-3x^{2}-4x+1$ to find the $y$-coordinate of this point

$\begin{gathered} y = 3^{3} - 3 \times 3^{2} - 4 \times 3 + 1 \\ = - 11 \end{gathered}$

Similarly, substitute $x=-1$ into the equation for $y$

$\begin{gathered} y = ( - 1 )^{3} - 3 \times ( - 1 )^{2} - 4 \times ( - 1 ) + 1 \\ = 1 \end{gathered}$

Write out the two sets of coordinates

Final answer: The points$(3,-11)$and$(-1,1)$have a gradient of 5