Differentiation

Multiply each term by its power of x, then reduce the power by 1; a number on its own differentiates to 0.

1. 2x^5 differentiates to 5\times2x^4=10x^4
2. -4x^3 differentiates to 3\times(-4)x^2=-12x^2
3. The constant 7 differentiates to 0
4. Answer: \dfrac{dy}{dx}=10x^4-12x^2

Write \dfrac{4}{x^3} as a power of x before you differentiate.

1. Write as powers of x: y=4x^{-3}+2x^5-1
2. 4x^{-3} differentiates to -3\times4x^{-4}=-12x^{-4} (one less than -3 is -4)
3. 2x^5 differentiates to 10x^4 and -1 to 0
4. Answer: \dfrac{dy}{dx}=10x^4-12x^{-4}

Expand the brackets first, then differentiate each term.

1. Expand: y=2x^3+6x-5x^2-15
2. Write in order: y=2x^3-5x^2+6x-15
3. Differentiate each term: 6x^2-10x+6 (the -15 differentiates to 0)
4. Answer: \dfrac{dy}{dx}=6x^2-10x+6

Do not differentiate the two factors separately and multiply: expand the bracket first.

1. Expand first: 4x^3\times x^2=4x^5 (add the powers)
2. and 4x^3\times(-2)=-8x^3, so y=4x^5-8x^3
3. Differentiate each term: 20x^4-24x^2
4. 24x^3 is 12x^2\times2x, the two factors differentiated separately and multiplied, which is wrong
5. Answer: 20x^4-24x^2

The rate of change of y with respect to x is \dfrac{dy}{dx}: differentiate, then substitute x=-2 using brackets.

1. Rate of change of y with respect to x means \dfrac{dy}{dx}
2. \dfrac{dy}{dx}=4x^3+9x^2-5
3. 4(-2)^3=-32 and 9(-2)^2=36
4. -32+36-5=-1
5. Answer: -1

Write y as a number times a power of x first: only the x^2 moves to the top as a negative power.

1. \dfrac{7}{3x^2}=\dfrac73\times\dfrac{1}{x^2}=\dfrac73x^{-2}
2. Multiply by the power, -2, and subtract 1 from it
3. \dfrac73\times(-2)x^{-3}=-\dfrac{14}{3}x^{-3}
4. -42x^{-3} comes from writing y as 21x^{-2}, which is wrong
5. Answer: -\dfrac{14}{3}x^{-3}

The gradient of the tangent at a point is the value of \dfrac{dy}{dx} there.

1. Write as powers of x: y=x^3-6x^{-1}
2. \dfrac{dy}{dx}=3x^2+6x^{-2}=3x^2+\dfrac{6}{x^2}
3. At x=3: 3\times9+\dfrac69=27+\dfrac23
4. Answer: \dfrac{83}{3} (or 27\frac23)

The gradient is \dfrac{dy}{dx}, not y: differentiate, then set the result equal to 7.

1. \dfrac{dy}{dx}=3x^2-8x+3
2. The gradient is 7: 3x^2-8x+3=7
3. Subtract 7 from both sides: 3x^2-8x-4=0
4. Putting y=7 gives x^3-4x^2+3x=0, which finds where the curve is at height 7, not where its gradient is 7
5. Answer: 3x^2-8x-4=0

Substitute x=-1 into \dfrac{dy}{dx}, not into y, taking care with odd and even powers of -1.

1. \dfrac{dy}{dx}=8x^3+3px^2-1
2. At x=-1: 8(-1)^3+3p(-1)^2-1=-8+3p-1
3. So 3p-9=-15
4. 3p=-6
5. Answer: p=-2

Expand the brackets first; notice that \dfrac{3}{x}\times2x is just a number.

1. x^2\times2x=2x^3 and x^2\times1=x^2
2. -\dfrac3x\times2x=-6 and -\dfrac3x\times1=-3x^{-1}
3. So y=2x^3+x^2-6-3x^{-1}
4. Differentiate each term: 6x^2+2x+3x^{-2}
5. Answer: \dfrac{dy}{dx}=6x^2+2x+\dfrac{3}{x^2}

Divide each term on the top by x^3 before you differentiate; do not differentiate the top and bottom separately.

1. Divide each term by x^3 (subtract the powers)
2. y=5x-2x^{-1}+9x^{-3}
3. 5x gives 5; -2x^{-1} gives +2x^{-2}; 9x^{-3} gives -27x^{-4}
4. Answer: \dfrac{dy}{dx}=5+2x^{-2}-27x^{-4}

Differentiate, set \dfrac{dy}{dx} equal to -3, then rearrange to find x^3.

1. Write as powers of x: y=5x-32x^{-2}
2. \dfrac{dy}{dx}=5+64x^{-3}=5+\dfrac{64}{x^3}
3. 5+\dfrac{64}{x^3}=-3, so \dfrac{64}{x^3}=-8
4. x^3=64\div(-8)=-8
5. Answer: x=-2

Write \dfrac{8}{x^2} as 8x^{-2}, differentiate twice, then substitute x=2

1. Write as powers of x: y=3x^4-5x+8x^{-2}
2. Differentiate: \dfrac{dy}{dx}=12x^3-5-16x^{-3}
3. Differentiate again: \dfrac{d^2y}{dx^2}=36x^2+48x^{-4}
4. Substitute x=2: 36\times4+\dfrac{48}{16}=144+3
5. Answer: 147

Write \dfrac{2}{3x^3} as \dfrac23x^{-3}, then differentiate twice.

1. Write as powers of x: y=5x^2-\dfrac23x^{-3}
2. \dfrac{dy}{dx}=10x+2x^{-4} (since -\dfrac23\times-3=2)
3. \dfrac{d^2y}{dx^2}=10+2\times(-4)x^{-5}
4. Answer: \dfrac{d^2y}{dx^2}=10-8x^{-5}

The rate of change of the gradient is \dfrac{d^2y}{dx^2}, so differentiate twice.

1. \dfrac{dy}{dx}=4x^3-6x^2+5
2. \dfrac{d^2y}{dx^2}=12x^2-12x
3. 12x^2-12x=24, so x^2-x-2=0
4. (x-2)(x+1)=0, so x=2 or x=-1
5. x is positive
6. Answer: x=2

Do not differentiate the top and bottom separately: expand and split the fraction into separate powers of x first.

Part (a)

1. Expand the top: (x-4)^2=x^2-8x+16
2. Divide each term by 2x: y=\dfrac{x}{2}-4+\dfrac{8}{x}
3. Write as powers of x: y=\frac12x-4+8x^{-1}
4. Differentiate: \frac12-8x^{-2}
5. Answer: \dfrac{dy}{dx}=\dfrac12-\dfrac{8}{x^2}

Part (b)

1. Rate of change of y with respect to x means \dfrac{dy}{dx}
2. \dfrac12-\dfrac{8}{x^2}=-\dfrac32
3. \dfrac{8}{x^2}=2, so x^2=4
4. Answer: x=2 or x=-2

Find the coordinates of A and B to get the gradient of AB, then set \dfrac{dy}{dx} equal to it. Answer part (a) before attempting part (b).

Part (a)

1. At x=-3: y=-27+9=-18, so A=(-3,\ -18)
2. At x=3: y=27-9=18, so B=(3,\ 18)
3. Gradient =\dfrac{18-(-18)}{3-(-3)}=\dfrac{36}{6}
4. Answer: 6

Part (b)

1. \dfrac{dy}{dx}=3x^2-3
2. Set it equal to the gradient of AB: 3x^2-3=6
3. 3x^2=9, so x^2=3
4. Both \sqrt3 and -\sqrt3 lie between -3 and 3
5. Answer: x=\sqrt3 and x=-\sqrt3

Use \dfrac{dy}{dx} at x=2 for one equation and \dfrac{d^2y}{dx^2} at x=2 for the other, then solve them simultaneously. Answer part (a) before attempting part (b).

Part (a)

1. y=px^3+qx^{-1}, so \dfrac{dy}{dx}=3px^2-qx^{-2}
2. At x=2: 12p-\dfrac q4=10
3. \dfrac{d^2y}{dx^2}=6px+2qx^{-3}
4. At x=2: 12p+\dfrac q4=14
5. Add the two equations: 24p=24
6. Answer: p=1

Part (b)

1. Substitute p=1 into 12p+\dfrac q4=14
2. \dfrac q4=2
3. Answer: q=8

Write \dfrac{2}{x} as 2x^{-1} before differentiating, and use (\sqrt2)^2=2 when you substitute.

Part (a)

1. y=x^4-3x^2+2x^{-1}, so \dfrac{dy}{dx}=4x^3-6x-2x^{-2}
2. (\sqrt2)^2=2 and (\sqrt2)^3=2\sqrt2
3. 4\times2\sqrt2-6\sqrt2-\dfrac22
4. =8\sqrt2-6\sqrt2-1
5. Answer: -1+2\sqrt2

Part (b)

1. \dfrac{d^2y}{dx^2}=12x^2-6+4x^{-3}
2. At x=\sqrt2: 12\times2-6+\dfrac{4}{2\sqrt2}
3. \dfrac{4}{2\sqrt2}=\dfrac{2}{\sqrt2}=\dfrac{2\sqrt2}{2}=\sqrt2
4. Answer: 18+\sqrt2

Set \dfrac{dy}{dx} equal to 1, divide through by 4 and use the factor theorem to find one factor of the cubic. Answer part (a) before attempting part (b).

Part (a)

1. Differentiate each term
2. Answer: \dfrac{dy}{dx}=4x^3-12x^2-4x+13

Part (b)

1. 4x^3-12x^2-4x+13=1
2. 4x^3-12x^2-4x+12=0
3. Divide by 4: x^3-3x^2-x+3=0
4. When x=1: 1-3-1+3=0, so (x-1) is a factor
5. x^3-3x^2-x+3=(x-1)(x^2-2x-3)=(x-1)(x-3)(x+1)
6. Answer: x=-1, x=1 and x=3