Differentiation Rules are formulas that allow us to take a complicated derivative and break it up into smaller pieces so we can apply more basic differentiation rules.

The product rule states that for the functions $u$ and $v$:

$\frac{d}{dx} \left(u\cdot v\right) = u\cdot\frac{dv}{dx} + v\frac{du}{dx}$

For example:

## Let $u = x^2$ and $v = e^x$. If $f(x) = u\cdot v$, find $f'(x)$.

$\begin{aligned} \frac{d}{dx}\left(x^2 \cdot e^x \right) &= \underbrace{x^2}_u \cdot \underbrace{e^x}_{\frac{dv}{dx}} + \underbrace{e^x}_v \cdot \underbrace{2x}_\frac{du}{dx} \\ &= e^x\left(x^2+2x\right)\, _\square \end{aligned}$

The quotient rule is useful when you have a rational expression. In Lagrange's notation:

$\left(\frac{p}{q}\right)' = \frac{q \cdot p' - p\cdot q'}{q^2}$

## Find $f'(x)$ if $f(x)=\frac{\sin x}{x^3}$.

Let $p=\sin x$ and let $q=x^3$. Thus, $p'=\cos x$ and $q'=3x^2$, so:

$\begin{aligned} f'(x)&=\frac{\left(x^3\right)\left( \cos x \right) - \left(\sin x\right)\left(3x^2\right)}{x^6} \\ &= \frac{x^2\left(x \cos x - 3 \sin x\right)}{x^6} \\ &= \frac{x \cos x - 3 \sin x}{x^4} \, _\square \end{aligned}$

The Chain Rule is needed for composite functions. Consider $f(x)=g(h(x)).$

The chain rule explains how we can relate $f'(x)$ to the derivative of $g$ and $h$. Specifically:

$f'(x) = g'(h(x)) \cdot h'(x).$

## Find the derivative of $f(x) = \sqrt{\cos x}$.

Let $g(x) = \sqrt{x}$, and $h(x) = \cos x$. Taking the derivatives of these functions, we get:

$g'(x)=\frac{1}{2 \sqrt{x}}$ $h'(x) = -\sin x$ So, according to the chain rule:

$\begin{aligned} f'(x) &= \underbrace{\frac{1}{2\sqrt{\cos x}}}_{g'(h(x))} \cdot \underbrace{-\sin x}_{h'(x)} \\ &= -\frac{\sin x}{2\sqrt{\cos x}} \, _\square \end{aligned}$

Let's consider a more complicated example:

## Find the derivative of $\displaystyle f(x) = \frac{\sqrt{x^2+1}}{x^2+3}$.

First, notice that this is rational function, so we'll need to use the quotient rule. The derivative of the denominator ( $x^2+3$ ) is simply $2x$. In order to use the quotient rule, however, we'll also need to know the derivative of the numerator, which we can't find directly.

In order to find the derivative of the numerator, $\sqrt{x^2+1}$, we'll use the chain rule: $\begin{aligned} \frac{d}{dx}\left( \sqrt{x^2+1} \right) &= \frac{1}{2\sqrt{x^2+1}}\cdot \frac{d}{dx}\left( x^2 + 1 \right) \\ &= \frac{1}{2\sqrt{x^2+1}}\cdot 2x \\ &= \frac{x}{\sqrt{x^2+1}} \end{aligned}$ Now that we know the derivatives of both the numerator and the denominator, we can proceed to use the quotient rule.

$f'(x) = \frac{\left(x^2+3\right)\left( \dfrac{x}{\sqrt{x^2+1}} \right) - \sqrt{x^2+1}\left(2x\right)}{\left( x^2 +3 \right)^2}$

Go through each term, and make sure you understand why it's there. You may want to refer again to the statement of the quotient rule above.

Now we should simplify, so factoring out $\frac{1}{\sqrt{x^2+1}}$, we get:

$f'(x) = \frac{1}{\sqrt{x^2+1}} \left(\frac{\left(x^2+3\right)\left( x \right) - \left(x^2+1\right) \left(2x\right)}{\left( x^2 +3 \right)^2}\right)$

Multiply and combine like terms:

$\begin{aligned} f'(x) &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x^3+3x-2x^3-2x}{\left( x^2 +3 \right)^2}\right) \\ &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x-x^3}{\left( x^2 +3 \right)^2}\right) \, _\square \end{aligned}$

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## Comments

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TopNewestplease differentiate e^x

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d(e^x)/dx=e^x

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differentiation of $e^{x}$ is $e^{x}$ only . we can do it by log method too. let y= $e^{x}$ ..........> Eq1 taking log both sides we get log y = log $e^{x}$ since log and exponential are inversely propotional log$e^{x}$is equal to x then by differentiating both sides we get ,

$\frac{1}{y}$ . $\frac{dy}{dx}$ = 1 ( w.k.t $\frac{d log}{dx}$ = $\frac{1}{x}$ ) \ (\frac{dy}{dx}) = $\frac{1}{y}$ . 1 =$e^{x}$ .1 = $e^{x}$ ( from eq 1 ) . hence differentiated hope u got your answer

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We can use the fact that: $\int \frac{1}{x} = Lnx (+c)$ to solve the following differential equation: $\frac{dy}{dx} = y$ $\Rightarrow\int \frac{1}{y} dy = \int 1 dx \Rightarrow\ Ln y = x +c$ $\therefore\ y = e^{x+c}$ $\therefore\ y= e^x \implies\frac{dy}{dx} = e^x$

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nicely explained!

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