Differentiate \( y = 3x^4 - 2x + 5 \) with respect to \( x \).
\( \dfrac{dy}{dx} = 12x^3 - 2 \).
Differentiation
12 exam-style questions from first principles to applications (gradients, stationary points, rates). Tap Show answer for full working.
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Find the gradient of the curve \( y = x^2 \) at the point where \( x = 3 \).
\( \dfrac{dy}{dx} = 2x \). At \( x=3 \), gradient \( = 6 \).
Differentiate \( y = \dfrac{1}{x} + \sqrt{x} \).
\( y = x^{-1} + x^{1/2} \). \( \dfrac{dy}{dx} = -x^{-2} + \dfrac{1}{2}x^{-1/2} = -\dfrac{1}{x^2} + \dfrac{1}{2\sqrt{x}} \).
Find the equation of the tangent to \( y = x^3 \) at \( x = 1 \).
\( \dfrac{dy}{dx} = 3x^2 \). At \( x=1 \), gradient 3, point \( (1,1) \). \( y-1 = 3(x-1) \) ⇒ \( y = 3x - 2 \).
Find the stationary points of \( y = x^3 - 3x + 2 \) and determine their nature.
\( \dfrac{dy}{dx} = 3x^2 - 3 = 0 \) ⇒ \( x^2=1 \) ⇒ \( x=\pm 1 \). \( \dfrac{d^2y}{dx^2} = 6x \). At \( x=1 \): second deriv positive → minimum \( (1,0) \). At \( x=-1 \): second deriv negative → maximum \( (-1,4) \).
Differentiate \( y = (2x+1)^5 \) using the chain rule.
\( \dfrac{dy}{dx} = 5(2x+1)^4 \cdot 2 = 10(2x+1)^4 \).
A curve has equation \( y = x^2 - 4x + 3 \). Find the coordinates of the point on the curve where the gradient is 2.
\( \dfrac{dy}{dx} = 2x - 4 = 2 \) ⇒ \( x=3 \). \( y=9-12+3=0 \). Point \( (3,0) \).
Find \( \dfrac{dy}{dx} \) when \( y = \dfrac{x^2+1}{x} \).
\( y = x + x^{-1} \). \( \dfrac{dy}{dx} = 1 - x^{-2} = 1 - \dfrac{1}{x^2} \).
From first principles, find the derivative of \( y = x^2 \).
\( \dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{(x+h)^2 - x^2}{h} = \lim_{h\to 0} \dfrac{2xh + h^2}{h} = \lim_{h\to 0}(2x+h) = 2x \).
The volume of a sphere is \( V = \dfrac{4}{3}\pi r^3 \). Find \( \dfrac{dV}{dr} \) and the rate of change of volume when \( r=5 \) cm and \( \dfrac{dr}{dt} = 0.1 \) cm/s.
\( \dfrac{dV}{dr} = 4\pi r^2 \). \( \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} = 4\pi\cdot 25\cdot 0.1 = 10\pi \) cm³/s.
Find the equation of the normal to the curve \( y = \dfrac{1}{x} \) at the point where \( x = 2 \).
\( \dfrac{dy}{dx} = -1/x^2 \). At \( x=2 \), gradient \( -1/4 \). Normal gradient \( 4 \). Point \( (2,1/2) \). \( y - 1/2 = 4(x-2) \) ⇒ \( y = 4x - 8 + 1/2 \) ⇒ \( y = 4x - 15/2 \).
A curve has equation \( y = x^3 - 6x^2 + 9x + 1 \). Find the coordinates of the stationary points and determine whether each is a maximum or minimum.
\( \dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)=0 \). \( x=1 \) or \( 3 \). \( y(1)=1-6+9+1=5 \); \( y(3)=27-54+27+1=1 \). Second deriv \( 6x-12 \): at \( x=1 \) negative → max \( (1,5) \); at \( x=3 \) positive → min \( (3,1) \).