Examples of calculating the derivative
Example 1: Find the derivative of the functions.
Applying the constant rule:
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}}}}$
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$ *
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.*
TIP! See also Examples with Limits
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 rule, power 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$
