Differentiation Rules - Practice Questions & MCQ

Rules of Differentiation (Sum/Difference/Product)

Let f(x) and g(x) be differentiable functions and k as constant. Then each of the following rules holds:

Sum Rule

The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g.

\begin{aligned}
&\frac{d}{d x}(f(x)+g(x))=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\\
&\text { In general, }\\
&\frac{d}{d x}(f(x)+g(x)+h(x)+\ldots \ldots)=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))+\frac{d}{d x}(h(x))+\ldots \ldots
\end{aligned}

Difference Rule

The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g.

\begin{aligned}
& \frac{d}{d x}(f(x)-g(x))=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x)) \\
& \frac{d}{d x}(f(x)-g(x)-h(x)-\ldots \ldots)=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x))-\frac{d}{d x}(h(x))-\ldots \ldots
\end{aligned}

Constant Multiple Rule

The derivative of a constant k multiplied by a function f is the same as the constant multiplied by the derivative of f

$
\frac{d}{d x}(k f(x))=k \frac{d}{d x}(f(x))
$
Product rule
Let $f(x)$ and $g(x)$ be differentiable functions. Then,

$
\frac{d}{d x}(f(x) g(x))=g(x) \cdot \frac{d}{d x}(f(x))+f(x) \cdot \frac{d}{d x}(g(x))
$

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Extending the Product Rule

If 3 functions are involved, i.e let $k(x)=f(x) \cdot g(x) \cdot h(x)$
Let us have a function $k(x)$ as the product of the function $f(x) g(x)$ and the function $h(x)$. That is, $k(x)=$ $(f(x) \cdot g(x)) \cdot h(x)$. Thus,

$
k^{\prime}(x)=\frac{d}{d x}(f(x) g(x)) \cdot h(x)+\frac{d}{d x}(h(x)) \cdot(f(x) g(x))
$

[By applying the product rule to the product of $f(x) g(x)$ and $h(x)$.]

$
\begin{aligned}
& =\left(f^{\prime}(x) g(x)+g^{\prime}(x) f(x)\right) h(x)+h^{\prime}(x) f(x) g(x) \\
& =f^{\prime}(x) g(x) h(x)+f(x) g^{\prime}(x) h(x)+f(x) g(x) h^{\prime}(x)
\end{aligned}
$

Rules of Differentiation (Division or Quotient Rule)
Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$
\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot \frac{d}{d x}(f(x))-f(x) \cdot \frac{d}{d x}(g(x))}{(g(x))^2}
$
OR
if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{(g(x))^2}$

As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives.

If $u(x)$ and $v(x)$ are differentiable functions, then $u o v(x)$ or $u[v(x)]$ is also differentiable.
If $y=\operatorname{uov}(x)=u[v(x)]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} u\{v(x)\}}{\mathrm{d}\{v(x)\}} \times \frac{\mathrm{d}}{\mathrm{~d} x} v(x)
$

Is known as the chain rule. Or,
If $y=f(u)$ and $u=g(x)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{~d} u}{\mathrm{~d} x}$
The chain rule can be extended as follows:
If $y=[\operatorname{uovow}(x)]=u[v\{w(x)\}]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d} v\{w(x)\}} \times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d} w(x)} \times \frac{\mathrm{d}[w(x)]}{\mathrm{d} x}
$