Find Derivatives of Functions in Calculus
Find the derivatives of various functions using different methods and rules in calculus. Several Examples with detailed solutions are presented. More exercises with answers are at the end of this page.
Example 1:
Find the derivative of the function \( f \) given by
\[
f(x) = (x^2 - 5)(x^3 - 2x + 3)
\]
Solution to Example 1:
Function \( f \) is the product of two functions:
\[
U = x^2 - 5 \quad \text{and} \quad V = x^3 - 2x + 3
\]
Hence,
\[
f(x) = U \, V
\]
We use the product rule to differentiate \( f \) as follows:
\[
f'(x) = U'V + UV'
\]
where \( U' \) and \( V' \) are the derivatives of \( U \) and \( V \) respectively, and are given by
\[
U' = 2x \quad \text{and} \quad V' = 3x^2 - 2
\]
Substitute to obtain:
\[
f'(x) = 2x(x^3 - 2x + 3) + (x^2 - 5)(3x^2 - 2)
\]
Expand, group, and simplify to obtain:
\[
\begin{align*}
f'(x) &= 2x^4 - 4x^2 + 6x + 3x^4 - 15x^2 + 10 \\
&= 5x^4 - 21x^2 + 6x + 10
\end{align*}
\]
Example 2:
Calculate the first derivative of function f given by
\[
f(x) = (\sqrt x + 2x)(4x^2-1)
\]
Solution to Example 2:
This function may be considered as the product of function \( U = \sqrt x + 2x \) and \( V = 4 x^2 - 1\), hence the use of the product rule
\[
f'(x) = U' V + U V' \\
= (\dfrac{1}{2\sqrt x} + 2)(4x^2-1) + (\sqrt x + 2 x)(8x)
\]
To add the above you need to write all terms as fractions with a common denominator.
\[
f'(x) = \dfrac{(1+2\cdot2\sqrt x)(4x^2-1)+2\sqrt x(8x)(\sqrt x + 2x)}{2\sqrt x}
\]
Expand
\[
f'(x) = \dfrac{4x^2-1+16x^{5/2}-4\sqrt x+16x^2+32x^{5/2}}{2\sqrt x}
\]
and group to obtain the final result for the derivative of f as follows.
\[
f'(x) = \dfrac{48x^{5/2}+20x^2-4x^{1/2}-1}{2\sqrt x}
\]
Example 3:
Calculate the first derivative of the function \( f \) given by
\[
f(x) = \dfrac{x^2 + 1}{5x - 3}
\]
Solution to Example 3:
The given function may be considered as the ratio of two functions:
\[
U = x^2 + 1 \quad \text{and} \quad V = 5x - 3
\]
We use the quotient rule to differentiate \( f \) as follows:
\[
f'(x) = \dfrac{U'V - UV'}{V^2}
\]
Since \( U' = 2x \) and \( V' = 5 \), we have:
\[
f'(x) = \dfrac{2x(5x - 3) - (x^2 + 1)5}{(5x - 3)^2}
\]
Expand and group to obtain \( f'(x) \) as follows:
\[
\begin{align*}
f'(x) &= \dfrac{10x^2 - 6x - 5x^2 - 5}{(5x - 3)^2} \\
&= \dfrac{5x^2 - 6x - 5}{(5x - 3)^2}
\end{align*}
\]
Example 4:
Calculate the first derivative of function f given by
\[ f(x) = \dfrac{x - \sqrt{x}}{5x^2 - 3} \]
Solution to Example 4:
Function \( f \) is the quotient of two functions hence the use of the quotient rule:
\[ f'(x) = \dfrac{(1 - \dfrac{1}{2\sqrt{x}})(5x^2 - 3) - (x - \sqrt{x})10x}{(5x^2 - 3)^2} \]
Write all terms in the numerator so that they have the same denominator \( 2\sqrt{x} \):
\[ f'(x) = \dfrac{(2\sqrt{x} - 1)(5x^2 - 3) - 2\sqrt{x}(x - \sqrt{x})10x}{2\sqrt{x}(5x^2 - 3)^2} \]
Expand and group like terms to obtain \( f'(x) \):
\[ f'(x) = \dfrac{-10x^{5/2} + 15x^2 - 6\sqrt{x} + 3}{2\sqrt{x}(5x^2 - 3)^2} \]
Example 5:
Calculate the first derivative of function f given by
Calculate the first derivative of function \( f \) given by
\[
f(x) = \left( \dfrac{1}{x} - 3 \right) \dfrac{x^2 + 3}{2x - 1}
\]
Solution to Example 5:
Function \( f \) given above may be considered as the product of functions \( U = \dfrac{1}{x} - 3 \) and \( V = \dfrac{x^2 + 3}{2x - 1} \), and function \( V \) may be considered as the quotient of two functions \( x^2 + 3 \) and \( 2x - 1 \). We use the product rule for \( f \) and the quotient rule for \( V \) as follows.
Let \( U = \dfrac{1}{x} - 3 \) and \( V = \dfrac{x^2 + 3}{2x - 1} \).
Then \( f(x) = U \cdot V \), so by the product rule:
\[
f'(x) = U' \cdot V + U \cdot V'
\]
First, compute \( U' \):
\[
U = x^{-1} - 3 \quad \Rightarrow \quad U' = -x^{-2} = -\dfrac{1}{x^2}
\]
Next, compute \( V' \) using the quotient rule:
\[
V = \dfrac{x^2 + 3}{2x - 1} \quad \Rightarrow \quad V' = \dfrac{(2x)(2x - 1) - (x^2 + 3)(2)}{(2x - 1)^2}
\]
\[
V' = \dfrac{4x^2 - 2x - 2x^2 - 6}{(2x - 1)^2} = \dfrac{2x^2 - 2x - 6}{(2x - 1)^2}
\]
Now substitute into the product rule:
\[
f'(x) = \left(-\dfrac{1}{x^2}\right) \cdot \dfrac{x^2 + 3}{2x - 1} + \left(\dfrac{1}{x} - 3\right) \cdot \dfrac{2x^2 - 2x - 6}{(2x - 1)^2}
\]
Write with a common denominator \( x^2(2x - 1)^2 \):
\[
f'(x) = \dfrac{-(x^2 + 3)(2x - 1) + \left(\dfrac{1}{x} - 3\right)(2x^2 - 2x - 6) \cdot x^2}{x^2(2x - 1)^2}
\]
Simplify the second term: \( \left(\dfrac{1}{x} - 3\right) \cdot x^2 = x - 3x^2 \), so:
\[
f'(x) = \dfrac{-(x^2 + 3)(2x - 1) + (x - 3x^2)(2x^2 - 2x - 6)}{x^2(2x - 1)^2}
\]
Expand both terms in the numerator:
First term:
\[
-(x^2 + 3)(2x - 1) = -(2x^3 - x^2 + 6x - 3) = -2x^3 + x^2 - 6x + 3
\]
Second term:
\[
(x - 3x^2)(2x^2 - 2x - 6) = 2x^3 - 2x^2 - 6x - 6x^4 + 6x^3 + 18x^2
\]
\[
= -6x^4 + (2x^3 + 6x^3) + (-2x^2 + 18x^2) - 6x
\]
\[
= -6x^4 + 8x^3 + 16x^2 - 6x
\]
Add both parts:
\[
(-2x^3 + x^2 - 6x + 3) + (-6x^4 + 8x^3 + 16x^2 - 6x)
\]
\[
= -6x^4 + (-2x^3 + 8x^3) + (x^2 + 16x^2) + (-6x - 6x) + 3
\]
\[
= -6x^4 + 6x^3 + 17x^2 - 12x + 3
\]
Thus:
\[
f'(x) = \dfrac{-6x^4 + 6x^3 + 17x^2 - 12x + 3}{x^2(2x - 1)^2}
\]
Example 6:
Calculate the first derivative of function \( f \) given by
\[
f(x) = \sqrt{x} (2x - 1)(x^3 - x)
\]
Solution to Example 6:
There are several ways to find the derivative of function \( f \) given above. One of them is to consider function \( f \) as the product of function \( U = \sqrt{x} \) and \( V = (2x - 1)(x^3 - x) \), and also consider \( V \) as the product of \( (2x - 1) \) and \( (x^3 - x) \), and apply the product rule to \( f \) and \( V \) as follows.
Let \( U = \sqrt{x} \) and \( V = (2x - 1)(x^3 - x) \).
Then \( f(x) = U \cdot V \), so by the product rule:
\[
f'(x) = U' \cdot V + U \cdot V'
\]
First, compute \( U' \):
\[
U = x^{1/2} \quad \Rightarrow \quad U' = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}
\]
Now compute \( V' \), where \( V = (2x - 1)(x^3 - x) \).
Let \( P = 2x - 1 \) and \( Q = x^3 - x \), then \( V' = P' \cdot Q + P \cdot Q' \):
\[
P' = 2, \quad Q' = 3x^2 - 1
\]
\[
V' = 2(x^3 - x) + (2x - 1)(3x^2 - 1)
\]
Now substitute into the product rule:
\[
f'(x) = \dfrac{1}{2\sqrt{x}} \cdot (2x - 1)(x^3 - x) + \sqrt{x} \cdot \left[2(x^3 - x) + (2x - 1)(3x^2 - 1)\right]
\]
Write with a common denominator \( 2\sqrt{x} \):
\[
f'(x) = \dfrac{(2x - 1)(x^3 - x) + 2x\left[2(x^3 - x) + (2x - 1)(3x^2 - 1)\right]}{2\sqrt{x}}
\]
Expand each term:
**First term:** \( (2x - 1)(x^3 - x) = 2x^4 - 2x^2 - x^3 + x \)
**Second term inside brackets:**
\[
2(x^3 - x) = 2x^3 - 2x
\]
\[
(2x - 1)(3x^2 - 1) = 6x^3 - 2x - 3x^2 + 1
\]
Add them: \( (2x^3 - 2x) + (6x^3 - 2x - 3x^2 + 1) = 8x^3 - 3x^2 - 4x + 1 \)
Multiply by \( 2x \):
\[
2x(8x^3 - 3x^2 - 4x + 1) = 16x^4 - 6x^3 - 8x^2 + 2x
\]
Now combine with first term:
\[
(2x^4 - 2x^2 - x^3 + x) + (16x^4 - 6x^3 - 8x^2 + 2x)
\]
\[
= (2x^4 + 16x^4) + (-x^3 - 6x^3) + (-2x^2 - 8x^2) + (x + 2x)
\]
\[
= 18x^4 - 7x^3 - 10x^2 + 3x
\]
Thus:
\[
f'(x) = \dfrac{18x^4 - 7x^3 - 10x^2 + 3x}{2\sqrt{x}}
\]
Exercises:
Find the derivative of each of the following functions.
- 1. \( f(x) = (x^3 + 2x)\sqrt{x} \)
- 2. \( g(x) = \dfrac{-(x+1)}{4x+7} \)
- 3. \( h(x) = (x+1)(\sqrt{x}-1)\left(\dfrac{1}{x}+4\right) \)
- 4. \( p(x) = \dfrac{\sqrt{x}+2x}{x^2-1} \)
- 5. \( q(x) = \sqrt{2x^4+2x-1} \)
- 6. \( r(x) = \dfrac{\sqrt{3x-1}}{-x^2+3} \)
Answers to Above Exercises:
- 1. \( f'(x) = \dfrac{\sqrt{x}(7x^2+6)}{2} \)
- 2. \( g'(x) = \dfrac{-3}{(4x+7)^2} \)
- 3. \( h'(x) = \dfrac{12x^{5/2} - 8x^2 + 5x^{3/2} - \sqrt{x} + 2}{2x^2} \)
- 4. \( p'(x) = -\dfrac{4x^{5/2} + 3x^2 + 4\sqrt{x} + 1}{2\sqrt{x}(x^2-1)^2} \)
- 5. \( q'(x) = \dfrac{4x^3 + 1}{\sqrt{2x^4 + 2x - 1}} \)
- 6. \( r'(x) = \dfrac{9x^2 - 4x + 9}{2(x^2 - 3)^2\sqrt{3x - 1}} \)
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