Coaching Centre
Differentiating powers with negative indices
Extending the power rule to negative indices, rewriting fractions as powers, and combining negative indices with the chain rule.
The power rule still works
The rule $\dfrac{d}{dx}\,x^n = nx^{n-1}$ holds for negative $n$ too. Rewrite first, differentiate, then write the answer back as a fraction.
Differentiate $y = \dfrac{3}{x^2}$.
Differentiate $y = \dfrac{1}{2x + 1}$.
Find the tangent to $y = \dfrac1x$ at $x = 2$.
The gradient is always negative. It gets steeper as the point moves towards the y-axis.
Where does each tangent cut the axes? Try a few values of $a$.
$\dfrac{d}{dx}\left(\dfrac1x\right) = $
$x^{-1} \to -x^{-2}$.
For $y = \dfrac{3}{x^2}$, find $\dfrac{dy}{dx}$ at $x = 1$.
$-\dfrac{6}{x^3} = -6$.
$\dfrac{d}{dx}\left(\dfrac{1}{2x + 1}\right) = $
Chain rule gives the extra factor $2$.
The tangent to $y = \dfrac1x$ at $x = 2$ is $y = -\tfrac14 x + b$. Find $b$.
$\tfrac12 + \tfrac24 = 1$.
For $y = \dfrac4x$, find the gradient at $x = -2$.
$-\dfrac{4}{x^2} = -1$.
Find the positive $x$ where $y = x + \dfrac1x$ has gradient $0$.
$1 - \dfrac{1}{x^2} = 0$, so $x = 1$.
$\dfrac{d}{dx}\left(5x^{-3}\right) = $
Multiply by $-3$, reduce the index to $-4$.
For $y = \dfrac{2}{(x - 3)^2}$, find $\dfrac{dy}{dx}$ at $x = 4$.
$-\dfrac{4}{(x - 3)^3} = -4$.
For $f(x) = \dfrac{x^2 + 1}{x}$, find $f'(2)$.
$f(x) = x + x^{-1}$, so $f'(x) = 1 - \dfrac{1}{x^2}$ and $f'(2) = \tfrac34$.
New numbers every time. Keep going until you get five in a row.