A rule for differentiating powers of x
The power rule, differentiating sums and multiples, expanding products first, and finding tangents and normals to a curve.
The power rule
Constant multiples and sums differentiate term by term: if $f(x) = ag(x) + bh(x)$, then $f'(x) = ag'(x) + bh'(x)$.
Differentiate $f(x) = 4x^3 - 5x^2 + 7x - 2$.
Differentiate $y = (2x + 1)(x - 3)$.
Tangents and normals
Find the tangent and normal to $y = x^2 - 4x + 5$ at $x = 3$.
Blue dashed: the tangent. Pink dashed: the normal, at right angles to it.
At which point is the normal vertical?
$\dfrac{d}{dx}\,x^7 = $
Bring down the power and reduce it by one.
For $f(x) = 4x^3 - 5x^2 + 7x - 2$, find $f'(1)$.
$12 - 10 + 7 = 9$.
The derivative of $(2x + 1)(x - 3)$ is:
Expand to $2x^2 - 5x - 3$ first.
Find the gradient of the tangent to $y = x^2 - 4x + 5$ at $x = 3$.
$2(3) - 4 = 2$.
The tangent to $y = x^2 - 4x + 5$ at $(3, 2)$ is:
$y - 2 = 2(x - 3)$.
The normal at that point has gradient:
Negative reciprocal of $2$.
Find $x$ where the tangent to $y = x^2 - 6x$ is horizontal.
$2x - 6 = 0$.
For $f(x) = \dfrac{x^3 + 2x^2}{x}$, find $f'(2)$.
$f(x) = x^2 + 2x$, so $f'(x) = 2x + 2$ and $f'(2) = 6$.
Find both $x$-values where $y = x^3 - 12x$ has gradient $0$.
$3x^2 - 12 = 0$, so $x = \pm 2$.
New numbers every time. Keep going until you get five in a row.
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