Category: Differentiation Rules

Constant Multiple Rule

The Constant Multiple Rule. If ${c}$ is a constant anf ${f{}}$ is a differentiable function then ${\left({c}{f{{\left({x}\right)}}}\right)}'={c}{\left({f{{\left({x}\right)}}}\right)}'$.

Proof. By definition ${\left({c}{f{{\left({x}\right)}}}\right)}'=\lim_{{{h}\to{0}}}\frac{{{c}{f{{\left({x}+{h}\right)}}}-{c}{f{{\left({x}\right)}}}}}{{h}}={c}\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={c}{f{'}}{\left({x}\right)}$.

Sum and Difference Rules

The Sum Rule. If $f$ and $g$ are both differentiable then ${\left({f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}\right)}'={\left({f{{\left({x}\right)}}}\right)}'+{\left({g{{\left({x}\right)}}}\right)}'$.

Product Rule

Product Rule. If $f$ and $g$ are both differentiable then ${\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}'={f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}+{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}$.

Quotient Rule

Quotient Rule. If $f$ and $g$ are differentiable, it can be stated that ${\left(\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right)}'=\frac{{{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}-{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}}}{{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}}$.

Chain Rule

Now, let's see how to differentiate composite functions.

Suppose that we are given a function ${h}{\left({x}\right)}={f{{\left({g{{\left({x}\right)}}}\right)}}}$. Remembering that ${g{'}}{\left({x}\right)}$ is the rate of change of ${g{{\left({x}\right)}}}$ with respect to ${x}$ and ${f{'}}{\left({g{{\left({x}\right)}}}\right)}$ is the rate of change of ${f{}}$ with respect to ${g{{\left({x}\right)}}}$, it is reasonable to suggest that the rate of change of ${f{}}$ with respect to ${x}$ is the product of ${f{'}}{\left({g{{\left({x}\right)}}}\right)}$ and ${g{'}}{\left({x}\right)}$.

Implicit Differentiation

The techniques we learned allow us to differentiate the functions like ${y}={\sqrt[{{3}}]{{{{x}}^{{2}}+{\sin{{\left({x}\right)}}}}}}$ or ${y}=\frac{{\tan{{\left({x}\right)}}}}{{{{e}}^{{{\sin{{\left({x}\right)}}}}}}}$, or, in general, ${y}={f{{\left({x}\right)}}}$. But what to do if we can't express ${y}$ in terms of ${x}$ or if the expression would be very complex?

Derivative of Inverse Function

Now, let's use implicit differentiation to find derivatives of inverse functions.

Fact. Suppose that the function ${y}={f{{\left({x}\right)}}}$ has unique inverse and has finite derivative ${f{'}}{\left({x}\right)}\ne{0}$ then the derivative of the inverse function ${y}={{f}}^{{-{1}}}{\left({x}\right)}$ is ${\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}$.

Logarithmic Differentiation

The calculation of the derivatives of functions involving products, powers, or quotients can be simplified with logarithmic differentiation (because of the properties of logarithms).

Let's see first how to differentiate the functions that already have a product and/or a quotient under the logarithm.