Chain rule

If $y = f (u)$ is differentiable at the point $u$ and $u = g (x)$ is differentiable at $x$ , then the composite function $(f \circ g) (x)$ is differentiable at $x$ and

{(f \circ g)}^{'} (x) = f^{'} (g (x)) g^{'} (x)

\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}

Examples

${[\frac{1}{{(3 x^{2} + 1)}^{2}}]}^{'} =$
${[{sin}^{3} (x)]}^{'} =$
${[{cos}^{4} (5 x^{3})]}^{'} =$
${[{(2 x^{3} + 2)}^{5} {(5 x + 8)}^{7}]}^{'} =$

${[k (f (g (x)))]}^{'} =$
${[{cos}^{3} (1 + \sqrt{x^{2} + \sqrt{x}})]}^{'} =$
${[sin {(cos \sqrt{2 + x} + cos \sqrt{2 + sin x})}^{2}]}^{'} =$
${[arcsin \sqrt{x - 1}]}^{'} =$
${[arcsin (x^{2} + 1)]}^{'} =$

Consider

f (x) = \sqrt{2 (x - 1)} - 1 g (x) = f^{-1} (x) = \frac{{(x + 1)}^{2}}{2} + 1

evaluate

g^{'} (1) = \frac{1}{f^{'} (g (1))} = \frac{1}{f^{'} (f^{-1} (1))} = \frac{1}{f^{'} (3)} = \frac{1}{2}

Inverse derivatives

Chain rule

f (g (x)) = x \Rightarrow f^{'} (g (x)) g^{'} (x) = 1 \Rightarrow g^{'} (x) = \frac{1}{f^{'} (g (x))}

{[f^{- 1}]}^{'} (x) = \frac{1}{f^{'} (f^{- 1} (x))}

Rational powers

Apply

f (g (x)) = x g^{'} (x) = \frac{1}{f^{'} (g (x))}

to rational powers

f (x) = x^{3} g (x) = \sqrt[3]{x}

f (x) = x^{n} g (x) = \sqrt[n]{x}

f (x) = x^{\frac{p}{q}} g (x) = \sqrt[\frac{q}{p}]{x}

Inverse trigonometric derivatives

f (g (x)) = x g^{'} (x) = \frac{1}{f^{'} (g (x))}

Apply to

f (x) = sin x g (x) = arcsin x