Calculus Examples

$y^{2} \ln (x) = 6$

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{2} \ln (x)) = \frac{d}{d x} (6)$

Differentiate the left side of the equation.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = y^{2}$ and $g (x) = \ln (x)$ .

$y^{2} \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [y^{2}]$

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

$y^{2} \frac{1}{x} + \ln (x) \frac{d}{d x} [y^{2}]$

Combine $y^{2}$ and $\frac{1}{x}$ .

$\frac{y^{2}}{x} + \ln (x) \frac{d}{d x} [y^{2}]$

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

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To apply the Chain Rule, set $u$ as $y$ .

$\frac{y^{2}}{x} + \ln (x) (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

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

$\frac{y^{2}}{x} + \ln (x) (2 u \frac{d}{d x} [y])$

Replace all occurrences of $u$ with $y$ .

$\frac{y^{2}}{x} + \ln (x) (2 y \frac{d}{d x} [y])$

Move $2$ to the left of $\ln (x)$ .

$\frac{y^{2}}{x} + 2 \cdot \ln (x) y \frac{d}{d x} [y]$

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{y^{2}}{x} + 2 \ln (x) y y'$

Reorder terms.

$2 y \ln (x) y' + \frac{y^{2}}{x}$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$0$

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

$2 y \ln (x) y' + \frac{y^{2}}{x} = 0$

Solve for $y'$ .

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Simplify $2 y \ln (x) y' + \frac{y^{2}}{x}$ .

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Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$y \ln (x^{2}) y' + \frac{y^{2}}{x} = 0$

Reorder factors in $y \ln (x^{2}) y' + \frac{y^{2}}{x}$ .

$y y' \ln (x^{2}) + \frac{y^{2}}{x} = 0$

Subtract $\frac{y^{2}}{x}$ from both sides of the equation.

$y y' \ln (x^{2}) = - \frac{y^{2}}{x}$

Divide each term in $y y' \ln (x^{2}) = - \frac{y^{2}}{x}$ by $y \ln (x^{2})$ and simplify.

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Divide each term in $y y' \ln (x^{2}) = - \frac{y^{2}}{x}$ by $y \ln (x^{2})$ .

$\frac{y y' \ln (x^{2})}{y \ln (x^{2})} = \frac{- \frac{y^{2}}{x}}{y \ln (x^{2})}$

Simplify the left side.

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Cancel the common factor of $y$ .

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Cancel the common factor.

$\frac{y y' \ln (x^{2})}{y \ln (x^{2})} = \frac{- \frac{y^{2}}{x}}{y \ln (x^{2})}$

Rewrite the expression.

$\frac{y' \ln (x^{2})}{\ln (x^{2})} = \frac{- \frac{y^{2}}{x}}{y \ln (x^{2})}$

Cancel the common factor of $\ln (x^{2})$ .

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Cancel the common factor.

$\frac{y' \ln (x^{2})}{\ln (x^{2})} = \frac{- \frac{y^{2}}{x}}{y \ln (x^{2})}$

Divide $y'$ by $1$ .

$y' = \frac{- \frac{y^{2}}{x}}{y \ln (x^{2})}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$y' = - \frac{y^{2}}{x} \cdot \frac{1}{y \ln (x^{2})}$

Cancel the common factor of $y$ .

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Move the leading negative in $- \frac{y^{2}}{x}$ into the numerator.

$y' = \frac{- y^{2}}{x} \cdot \frac{1}{y \ln (x^{2})}$

Factor $y$ out of $- y^{2}$ .

$y' = \frac{y (- y)}{x} \cdot \frac{1}{y \ln (x^{2})}$

Factor $y$ out of $y \ln (x^{2})$ .

$y' = \frac{y (- y)}{x} \cdot \frac{1}{y (\ln (x^{2}))}$

Cancel the common factor.

$y' = \frac{y (- y)}{x} \cdot \frac{1}{y \ln (x^{2})}$

Rewrite the expression.

$y' = \frac{- y}{x} \cdot \frac{1}{\ln (x^{2})}$

Multiply $\frac{- y}{x}$ by $\frac{1}{\ln (x^{2})}$ .

$y' = \frac{- y}{x \ln (x^{2})}$

Move the negative in front of the fraction.

$y' = - \frac{y}{x \ln (x^{2})}$

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

$\frac{d y}{d x} = - \frac{y}{x \ln (x^{2})}$