##### Examples of calculating the derivative

Example 1: Find the derivative of the functions.

Applying the

If $\displaystyle y=c$ then $\displaystyle y’=0$

If $\displaystyle y=a{{x}^{n}}$ then $\displaystyle y’=an{{x}^{{n-1}}}$

a) $\displaystyle y=3{{x}^{7}}$

$\displaystyle y’=(3\cdot 7){{x}^{{7-1}}}=21{{x}^{6}}$

b) $\displaystyle y=-4$

$\displaystyle y’=(-4)’=0$

c) $\displaystyle y=2{{x}^{{-1}}}$

$\displaystyle y’=\left[ {(-1)\cdot 2} \right]{{x}^{{(-1)-1}}}=-2{{x}^{{-2}}}$

d) $\displaystyle y=-{{x}^{{-\frac{1}{4}}}}$

$\displaystyle y’=(-1)\cdot (-\frac{1}{4}){{x}^{{(-\frac{1}{4}-1)}}}=\frac{1}{4}{{x}^{{-\frac{5}{4}}}}$

Example 2: Find the derivative of the functions. Firstly change the root term  into a power term and then apply the power function rule.

a) $\displaystyle y=\sqrt{x}$

$\displaystyle y={{x}^{{\frac{1}{2}}}}$

$\displaystyle y’=\frac{1}{2}{{x}^{{(\frac{1}{2}-1)}}}=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$

b) $\displaystyle y=\sqrt[3]{{{{x}^{2}}}}$

$\displaystyle y={{x}^{{\frac{2}{3}}}}$

$\displaystyle y’=\frac{2}{3}{{x}^{{(\frac{2}{3}-1)}}}=\frac{2}{3}{{x}^{{-\frac{1}{3}}}}$

c) $\displaystyle y=\frac{3}{{\sqrt{{{{x}^{3}}}}}}$

$\displaystyle y=3{{x}^{{\frac{3}{2}}}}$

$\displaystyle y’=3\cdot \frac{3}{2}{{x}^{{(\frac{3}{2}-1)}}}=\frac{9}{2}{{x}^{{\frac{1}{2}}}}$

Example 3: Find the derivative of the functions.

Applying the sum and difference rule.

$\displaystyle (f+g)’=f’+g’$ and $\displaystyle (f-g)’=f’-g’$

a) $\displaystyle y={{x}^{2}}+2x-3$

$\displaystyle y’=({{x}^{2}})’+(2x)’-(3)’=2x+2$

b) $\displaystyle y={{x}^{2}}-{{x}^{4}}+\pi$

$\displaystyle y’=({{x}^{2}})’-({{x}^{4}})’+(\pi )’=2x-4{{x}^{3}}$

Note: * $\displaystyle \pi$ is a constant since it’s a number $\displaystyle \pi \approx 3,14$ *

Example 4: Find the derivative of the functions. Applying the Product rule $\displaystyle (f\cdot g)’=f’g+g’f$

a) $\displaystyle y=({{x}^{2}}-5)(2x+7)$

$\displaystyle y’=2x(2x+7)+2({{x}^{2}}-5)$

$\displaystyle y’=4{{x}^{2}}+14x+2{{x}^{2}}-10$

$\displaystyle y’=6{{x}^{2}}+14x-10$

b) $\displaystyle y=(\sqrt{x}+3x)(5{{x}^{3}}+1)$

$\displaystyle y=({{x}^{{\frac{1}{2}}}}+3x)(5{{x}^{3}}+1)$

$\displaystyle y’=({{x}^{{\frac{1}{2}}}}+3x)'(5{{x}^{3}}+1)+(5{{x}^{3}}+1)'(\sqrt{x}+3x)$

$\displaystyle y’=({{x}^{{-\frac{1}{2}}}}+3)(5{{x}^{3}}+1)+15{{x}^{2}}(\sqrt{x}+3x)$

*Another way to find this derivative is by expanding the brackets and applying the sum or difference rule.*

a) $\displaystyle y=({{x}^{2}}-5)(2x+7)$

$\displaystyle y=({{x}^{2}})(2x)+7{{x}^{2}}+(-5)\cdot 2x+(-5)\cdot 7$

$\displaystyle y=2{{x}^{3}}+7{{x}^{2}}-10x-35$

$\displaystyle y’=(2{{x}^{3}})’+(7{{x}^{2}})’-(10x)’-(35)’$

$\displaystyle y’=6{{x}^{2}}+14x-10+0$

$\displaystyle y’=6{{x}^{2}}+14x-10$

Example 5: Find the derivatives of the functions.

Applying the Quotient rule

$\displaystyle {{\left( {\frac{f}{g}} \right)}^{‘}}=\frac{{f’g-g’f}}{{{{g}^{2}}}}$

a) $\displaystyle y=\frac{{2{{x}^{2}}+1}}{{x-1}}$

$\displaystyle y’=\frac{{(2{{x}^{2}}+1)'(x-1)-(x-1)'(2{{x}^{2}}+1)}}{{{{{(x-1)}}^{2}}}}$

$\displaystyle y’=\frac{{4x(x-1)-1(2{{x}^{2}}+1)}}{{{{{(x-1)}}^{2}}}}$

$\displaystyle y’=\frac{{4{{x}^{2}}-4x-2{{x}^{2}}-1}}{{{{{(x-1)}}^{2}}}}$

$\displaystyle y’=\frac{{2{{x}^{2}}-4x-1}}{{{{{(x-1)}}^{2}}}}$

b) $\displaystyle y=\frac{{x+\sqrt{x}}}{x}$

$\displaystyle y’=\frac{{(x+\sqrt{x})’x-x'(x+\sqrt{x})}}{{{{x}^{2}}}}$

$\displaystyle y’=\frac{{(1+{{x}^{{-\frac{1}{2}}}})x+1(x+\sqrt{x})}}{{{{x}^{2}}}}$

$\displaystyle y’=\frac{{2x+{{x}^{{(1-\frac{1}{2})}}}+\sqrt{x}}}{{{{x}^{2}}}}$

$\displaystyle y’=\frac{{2x+{{x}^{{\frac{1}{2}}}}+\sqrt{x}}}{{{{x}^{2}}}}$

$\displaystyle y’=\frac{{2x+\sqrt{x}+\sqrt{x}}}{{{{x}^{2}}}}$

$\displaystyle y’=\frac{{2x+2\sqrt{x}}}{{{{x}^{2}}}}$

*Another easy way to find the derivative of the second function is by simplyfing the expresion first and writing the terms as power terms.*

b) $\displaystyle y=\frac{{x+\sqrt{x}}}{x}$

Simply the expression first:

$\displaystyle y=\frac{x}{x}+\frac{{\sqrt{x}}}{x}$

$\displaystyle y=1+\frac{{{{x}^{{\frac{1}{2}}}}}}{x}$

$\displaystyle y=1+{{x}^{{(\frac{1}{2}-1)}}}$

$\displaystyle y=1+{{x}^{{-\frac{1}{2}}}}$

Find the derivative by applying the sum rulepower rule and constant rule:

$\displaystyle y’=(1)’+({{x}^{{-\frac{1}{2}}}})’$

$\displaystyle y’=0-\frac{1}{2}{{x}^{{(1-\frac{1}{2})}}}$

$\displaystyle y’=\frac{1}{2}{{x}^{{\frac{1}{2}}}}=\frac{1}{2}\sqrt{x}$

Example 6: Find the derivatives of the composite functions.

Applying the chain rule $\displaystyle (f\circ g)’=(f\circ g)’g’$

a) $\displaystyle y=\sqrt{{{{x}^{2}}+2x+3}}$

$\displaystyle y’={{\left( {\sqrt{{{{x}^{2}}+2x+3}}} \right)}^{‘}}({{x}^{2}}+2x+3)’$

$\displaystyle y’=\frac{1}{{2\sqrt{{{{x}^{2}}+2x+3}}}}\cdot (2x+2)$

$\displaystyle y’=\frac{{2(x+1)}}{{2\sqrt{{{{x}^{2}}+2x+3}}}}$

$\displaystyle y’=\frac{{(x+1)}}{{\sqrt{{{{x}^{2}}+2x+3}}}}$

b) $\displaystyle y={{(3{{x}^{2}}+2x)}^{2}}$

$\displaystyle y’={{\left[ {{{{(3{{x}^{2}}+2x)}}^{2}}} \right]}^{‘}}(3{{x}^{2}}+2x)’$

$\displaystyle y’=2(3{{x}^{2}}+2x)(6x+2)$

$\displaystyle y’=2(18{{x}^{3}}+6{{x}^{2}}+12{{x}^{2}}+4x)$

$\displaystyle y’=2(18{{x}^{3}}+18{{x}^{2}}+4x)$

$\displaystyle y’=36{{x}^{3}}+36{{x}^{2}}+8x$