Given $y = (1 + 2x)^6$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$12(1 + 2x)^5$
Given $y = (1 - x)^8$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$-8(1 - x)^7$
Given $y = (3 - x^2)^5$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$-10x(3 - x^2)^4$
Given $y = (2 - 3x^4)^4$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$-48x^3(2 - 3x^4)^3$
Given $\mathrm{f}(x) = (4 - 2x^2)^4$, find $\mathrm{f}'(x)$
$-16x(4 - 2x^2)^3$
Given $\mathrm{f}(x) = (x-3x^2)^{-2}$, find $\mathrm{f}'(x)$
$-2(1-6x)(x-3x^2)^{-3}$
Given $\mathrm{f}(x) = (4\sqrt{x} - 1)^4$, find $\mathrm{f}'(x)$
$8x^{-\frac{1}{2}}(4\sqrt{x} - 1)^3$
Given $\mathrm{f}(x) = (2x\sqrt{x} + 3x^3)^5$, find $\mathrm{f}'(x)$
$5(3\sqrt{x}+9x^2)(2x\sqrt{x} + 3x^3)^4$
Find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left( \sqrt{5x-2} \right)$
$\dfrac{5}{2}(5x-2)^{-\frac{1}{2}}$
Find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left( \dfrac{1}{2x-5} \right)$
$-2(2x-5)^{-2}$
Find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left( \sqrt{x-1}(x-1) \right)$
$\dfrac{3}{2}\sqrt{x-1}$
Find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left( \dfrac{3}{(2 - x)^5} \right)$
$15(2-x)^{-6}$
Find the equation of the tangent to the curve $$y = \dfrac{4}{2x+5}$$ at the point where $x = 0$.
$y = -\dfrac{8}{25}x + \dfrac{4}{5}$
Find the equation of the normal to the curve $$y = \dfrac{2}{(4-x)^2}$$ at the point where $x = 2$.
$y = -2x + 4.5$
The curve $$y = (2+3x)^3$$ has a stationary point. Use differentiation to find the coordinates of the stationary point and show that it is a point of inflection.
$\left(-\dfrac{2}{3},0\right)$
The curve $$y = (4-x)^4$$ has a stationary point. Use differentiation to find the coordinates of the stationary point and determine whether it is a maximum, minimum, or point of inflection.
$(4,0)$