Calculus Examples

$y = x + 3 \ln (5 x) - 4 x^{2} + e^{2 x} - π$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (x + 3 \ln (5 x) - 4 x^{2} + e^{2 x} - π)$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $x + 3 \ln (5 x) - 4 x^{2} + e^{2 x} - π$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3 \ln (5 x)] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3 \ln (5 x)] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [3 \ln (5 x)] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2

Evaluate $\frac{d}{d x} [3 \ln (5 x)]$ .

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Step 3.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \ln (5 x)$ with respect to $x$ is $3 \frac{d}{d x} [\ln (5 x)]$ .

$1 + 3 \frac{d}{d x} [\ln (5 x)] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 5 x$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $5 x$ .

$1 + 3 (\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.2.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

$1 + 3 (\frac{1}{u_{1}} \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $5 x$ .

$1 + 3 (\frac{1}{5 x} \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$1 + 3 (\frac{1}{5 x} (5 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + 3 (\frac{1}{5 x} (5 \cdot 1)) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.5

Multiply $5$ by $1$ .

$1 + 3 (\frac{1}{5 x} \cdot 5) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.6

Combine $\frac{1}{5 x}$ and $5$ .

$1 + 3 \frac{5}{5 x} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.7

Cancel the common factor of $5$ .

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Step 3.2.7.1

Cancel the common factor.

$1 + 3 \frac{5}{5 x} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.7.2

Rewrite the expression.

$1 + 3 \frac{1}{x} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.2.8

Combine $3$ and $\frac{1}{x}$ .

$1 + \frac{3}{x} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.3

Evaluate $\frac{d}{d x} [- 4 x^{2}]$ .

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Step 3.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{2}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{2}]$ .

$1 + \frac{3}{x} - 4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$1 + \frac{3}{x} - 4 (2 x) + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.3.3

Multiply $2$ by $- 4$ .

$1 + \frac{3}{x} - 8 x + \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- π]$

Step 3.4

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 3.4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 3.4.1.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

$1 + \frac{3}{x} - 8 x + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [2 x] + \frac{d}{d x} [- π]$

Step 3.4.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$1 + \frac{3}{x} - 8 x + e^{u_{2}} \frac{d}{d x} [2 x] + \frac{d}{d x} [- π]$

Step 3.4.1.3

Replace all occurrences of $u_{2}$ with $2 x$ .

$1 + \frac{3}{x} - 8 x + e^{2 x} \frac{d}{d x} [2 x] + \frac{d}{d x} [- π]$

Step 3.4.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$1 + \frac{3}{x} - 8 x + e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- π]$

Step 3.4.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{3}{x} - 8 x + e^{2 x} (2 \cdot 1) + \frac{d}{d x} [- π]$

Step 3.4.4

Multiply $2$ by $1$ .

$1 + \frac{3}{x} - 8 x + e^{2 x} \cdot 2 + \frac{d}{d x} [- π]$

Step 3.4.5

Move $2$ to the left of $e^{2 x}$ .

$1 + \frac{3}{x} - 8 x + 2 e^{2 x} + \frac{d}{d x} [- π]$

Step 3.5

Since $- π$ is constant with respect to $x$ , the derivative of $- π$ with respect to $x$ is $0$ .

$1 + \frac{3}{x} - 8 x + 2 e^{2 x} + 0$

Step 3.6

Simplify.

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Step 3.6.1

Add $1 + \frac{3}{x} - 8 x + 2 e^{2 x}$ and $0$ .

$1 + \frac{3}{x} - 8 x + 2 e^{2 x}$

Step 3.6.2

Reorder terms.

$- 8 x + \frac{3}{x} + 2 e^{2 x} + 1$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = - 8 x + \frac{3}{x} + 2 e^{2 x} + 1$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - 8 x + \frac{3}{x} + 2 e^{2 x} + 1$