Browse Differentiation

347 questions at your level

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Differentiation from first principles

8 questions

LessonNot started
Given that

y=5x2−3xy = 5x^2 - 3x

use differentiation from first principles to show that

dydx=10x−3\frac{dy}{dx} = 10x - 3

(3 marks)
●●●●●Level 23 marksStart

Differentiating powers of x

38 questions

LessonNot started
Given that

y=(2x+3)2x,x>0y = \dfrac{(2x + 3)^2}{\sqrt{x}}, \qquad x > 0

find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}, giving each term in its simplest form. (3 marks)
●●●●●Level 23 marksStart
f(x)=x3−3x2−24x+7\text{f}(x) = x^3 - 3x^2 - 24x + 7
(a) Find f′′(x)\text{f}''(x)

(2 marks)
(b) (i) Solve f′′(x)=0\text{f}''(x) = 0

(ii) Hence find the range of values of xx for which f\text{f} is concave.

(2 marks)
●●●●●Level 24 marksStart
Differentiate the following with respect to xx.
(a) 6x3+4x\dfrac{6}{x^3} + 4\sqrt{x}

(2 marks)
(b) x2e3xx^2e^{3x}

(2 marks)
(c) 2x+1x−3\dfrac{2x + 1}{x - 3}

(3 marks)
●●●●●Level 27 marksStart

Tangents and normals

55 questions

LessonNot started
Find an equation of the tangent to the curve

y=(x−3)5y = (x - 3)^5

at the point where x=0x = 0

(3 marks)
●●●●●Level 23 marksStart

Differentiating trigonometric functions

31 questions

LessonNot started
Differentiate y=sin⁡4x+cos⁡2xy = \sin 4x + \cos 2x. (2 marks)
●●●●●Level 22 marksStart
Differentiate y=cos⁡(5x)y = \cos(5x).
●●●●●Level 22 marksStart
Differentiate y=sin⁡(3x2)y = \sin(3x^2).
●●●●●Level 22 marksStart
Differentiate y=tan⁡(2x)y = \tan(2x).
●●●●●Level 22 marksStart
(a) Differentiate the following with respect to xx.

(i) ln⁡(3x2+5)\ln(3x^2 + 5)

(ii) xcos⁡4xx\cos 4x

(5 marks)
(b) Find ∫sin⁡3x dx\displaystyle\int \sin 3x\,dx.

(2 marks)
(c) The gradient of a curve is given by dydx=4e2x\dfrac{dy}{dx} = 4e^{2x} and the curve passes through the point (0,5)(0, 5).

Find the equation of the curve.

(2 marks)
●●●●●Level 29 marksStart
Differentiate y=sin⁡(x2+1)y = \sin(x^{2} + 1). (2 marks)
●●●●●Level 22 marksStart

Differentiating exponentials and logarithms

55 questions

LessonNot started
Differentiate y=e5x+e−2xy = \mathrm{e}^{5x} + \mathrm{e}^{-2x}. (2 marks)
●●●●●Level 22 marksStart
Differentiate y=ln⁡(x3)y = \ln(x^{3}), simplifying your answer. (2 marks)
●●●●●Level 22 marksStart
Differentiate y=e3xy = e^{3x}.
●●●●●Level 22 marksStart
Differentiate y=ln⁡(3x2+1)y = \ln(3x^2 + 1).
●●●●●Level 22 marksStart
Differentiate y=e2x2−1y = e^{2x^2 - 1}.
●●●●●Level 22 marksStart
A curve has equation y=6e−2xy = 6e^{-2x}

Find the gradient of the curve at the point where y=15y = 15

(3 marks)
●●●●●Level 23 marksStart
Differentiate (x2+4x−1)e−x(x^2 + 4x - 1)e^{-x} with respect to xx, giving your answer in the form e−x(ax2+bx+c)e^{-x}(ax^2 + bx + c), where aa, bb and cc are integers to be determined.

(3 marks)
●●●●●Level 23 marksStart
Differentiate the following with respect to xx.
(a) 6x3+4x\dfrac{6}{x^3} + 4\sqrt{x}

(2 marks)
(b) x2e3xx^2e^{3x}

(2 marks)
(c) 2x+1x−3\dfrac{2x + 1}{x - 3}

(3 marks)
●●●●●Level 27 marksStart
(a) Differentiate the following with respect to xx.

(i) ln⁡(3x2+5)\ln(3x^2 + 5)

(ii) xcos⁡4xx\cos 4x

(5 marks)
(b) Find ∫sin⁡3x dx\displaystyle\int \sin 3x\,dx.

(2 marks)
(c) The gradient of a curve is given by dydx=4e2x\dfrac{dy}{dx} = 4e^{2x} and the curve passes through the point (0,5)(0, 5).

Find the equation of the curve.

(2 marks)
●●●●●Level 29 marksStart

The chain rule

37 questions

LessonNot started
Differentiate y=(3x2−1)5y = (3x^2 - 1)^5.
●●●●●Level 22 marksStart
Differentiate y=(2x+5)4y = (2x + 5)^4.
●●●●●Level 22 marksStart
Differentiate y=(5x−2)3y = (5x - 2)^{3}. (2 marks)
●●●●●Level 22 marksStart
Find an equation of the tangent to the curve

y=(x−3)5y = (x - 3)^5

at the point where x=0x = 0

(3 marks)
●●●●●Level 23 marksStart
(a) Differentiate the following with respect to xx.

(i) ln⁡(3x2+5)\ln(3x^2 + 5)

(ii) xcos⁡4xx\cos 4x

(5 marks)
(b) Find ∫sin⁡3x dx\displaystyle\int \sin 3x\,dx.

(2 marks)
(c) The gradient of a curve is given by dydx=4e2x\dfrac{dy}{dx} = 4e^{2x} and the curve passes through the point (0,5)(0, 5).

Find the equation of the curve.

(2 marks)
●●●●●Level 29 marksStart

The product rule

40 questions

LessonNot started
Differentiate y=x2(2x+1)3y = x^2(2x + 1)^3, giving your answer in fully factorised form.
●●●●●Level 23 marksStart
Differentiate y=(x+1)(x−3)4y = (x + 1)(x - 3)^4, giving your answer in factorised form.
●●●●●Level 23 marksStart
Differentiate (x2+4x−1)e−x(x^2 + 4x - 1)e^{-x} with respect to xx, giving your answer in the form e−x(ax2+bx+c)e^{-x}(ax^2 + bx + c), where aa, bb and cc are integers to be determined.

(3 marks)
●●●●●Level 23 marksStart
Differentiate the following with respect to xx.
(a) 6x3+4x\dfrac{6}{x^3} + 4\sqrt{x}

(2 marks)
(b) x2e3xx^2e^{3x}

(2 marks)
(c) 2x+1x−3\dfrac{2x + 1}{x - 3}

(3 marks)
●●●●●Level 27 marksStart
(a) Differentiate the following with respect to xx.

(i) ln⁡(3x2+5)\ln(3x^2 + 5)

(ii) xcos⁡4xx\cos 4x

(5 marks)
(b) Find ∫sin⁡3x dx\displaystyle\int \sin 3x\,dx.

(2 marks)
(c) The gradient of a curve is given by dydx=4e2x\dfrac{dy}{dx} = 4e^{2x} and the curve passes through the point (0,5)(0, 5).

Find the equation of the curve.

(2 marks)
●●●●●Level 29 marksStart

The quotient rule

32 questions

LessonNot started
Differentiate y=xx+1y = \dfrac{x}{x + 1}.
●●●●●Level 22 marksStart
Differentiate y=3x−1x+2y = \dfrac{3x - 1}{x + 2}, simplifying your answer. (3 marks)
●●●●●Level 23 marksStart
Differentiate the following with respect to xx.
(a) 6x3+4x\dfrac{6}{x^3} + 4\sqrt{x}

(2 marks)
(b) x2e3xx^2e^{3x}

(2 marks)
(c) 2x+1x−3\dfrac{2x + 1}{x - 3}

(3 marks)
●●●●●Level 27 marksStart
The function f is defined by

f(x)=2x2+7x+3x2−x−12−6x−4x>4\text{f}(x) = \dfrac{2x^2 + 7x + 3}{x^2 - x - 12} - \dfrac{6}{x - 4} \qquad x > 4
(a) Show that

f(x)=2x−5x−4x>4\text{f}(x) = \dfrac{2x - 5}{x - 4} \qquad x > 4

(3 marks)
(b) Hence find f′(x)\text{f}'(x) giving your answer in simplest form.

(2 marks)
●●●●●Level 25 marksStart

Parametric differentiation

41 questions

LessonNot started
A curve has parametric equations x=t2, y=t3x = t^2,\ y = t^3. Find dydx\dfrac{dy}{dx} in terms of tt.
●●●●●Level 22 marksStart
A curve has parametric equations x=t2+1, y=t3−3tx = t^2 + 1,\ y = t^3 - 3t. Find dydx\dfrac{dy}{dx} in terms of tt, and its value at t=2t = 2.
●●●●●Level 23 marksStart
A curve has parametric equations x=4t,  y=t2−1x = 4t, \; y = t^{2} - 1.

Find dydx\dfrac{dy}{dx} in terms of tt. (2 marks)
●●●●●Level 22 marksStart

Convex, concave and points of inflection

37 questions

LessonNot started
Find the set of values of xx for which the curve y=x3−12xy = x^{3} - 12x is convex. (2 marks)
●●●●●Level 22 marksStart
Find the set of values of xx for which the curve y=x3−6x2+5y = x^3 - 6x^2 + 5 is convex.
●●●●●Level 23 marksStart
f(x)=x3−3x2−24x+7\text{f}(x) = x^3 - 3x^2 - 24x + 7
(a) Find f′′(x)\text{f}''(x)

(2 marks)
(b) (i) Solve f′′(x)=0\text{f}''(x) = 0

(ii) Hence find the range of values of xx for which f\text{f} is concave.

(2 marks)
●●●●●Level 24 marksStart

Connected rates of change

25 questions

LessonNot started
The radius of a circular oil slick increases at a rate of 0.2 m s−10.2\text{ m s}^{-1}. Find the rate at which its area is increasing at the instant when the radius is 10 m10\text{ m}.
●●●●●Level 23 marksStart