Calculus Examples

$y^{4} + x^{3} = y^{2} + 10 x$ , $(0, 1)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 1$ to find the slope of the tangent line.

Tap for more steps...

Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{4} + x^{3}) = \frac{d}{d x} (y^{2} + 10 x)$

Step 1.2

Differentiate the left side of the equation.

Tap for more steps...

Step 1.2.1

By the Sum Rule, the derivative of $y^{4} + x^{3}$ with respect to $x$ is $\frac{d}{d x} [y^{4}] + \frac{d}{d x} [x^{3}]$ .

$\frac{d}{d x} [y^{4}] + \frac{d}{d x} [x^{3}]$

Step 1.2.2

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

Tap for more steps...

Step 1.2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = y$ .

Tap for more steps...

Step 1.2.2.1.1

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

$\frac{d}{d u} [u^{4}] \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 1.2.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$4 u^{3} \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 1.2.2.1.3

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

$4 y^{3} \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 1.2.2.2

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

$4 y^{3} y' + \frac{d}{d x} [x^{3}]$

Step 1.2.3

Differentiate using the Power Rule.

Tap for more steps...

Step 1.2.3.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 y^{3} y' + 3 x^{2}$

Step 1.2.3.2

Reorder terms.

$3 x^{2} + 4 y^{3} y'$

Step 1.3

Differentiate the right side of the equation.

Tap for more steps...

Step 1.3.1

By the Sum Rule, the derivative of $y^{2} + 10 x$ with respect to $x$ is $\frac{d}{d x} [y^{2}] + \frac{d}{d x} [10 x]$ .

$\frac{d}{d x} [y^{2}] + \frac{d}{d x} [10 x]$

Step 1.3.2

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

Tap for more steps...

Step 1.3.2.1

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

Tap for more steps...

Step 1.3.2.1.1

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

$\frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [10 x]$

Step 1.3.2.1.2

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

$2 u \frac{d}{d x} [y] + \frac{d}{d x} [10 x]$

Step 1.3.2.1.3

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

$2 y \frac{d}{d x} [y] + \frac{d}{d x} [10 x]$

Step 1.3.2.2

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

$2 y y' + \frac{d}{d x} [10 x]$

Step 1.3.3

Evaluate $\frac{d}{d x} [10 x]$ .

Tap for more steps...

Step 1.3.3.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x$ with respect to $x$ is $10 \frac{d}{d x} [x]$ .

$2 y y' + 10 \frac{d}{d x} [x]$

Step 1.3.3.2

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

$2 y y' + 10 \cdot 1$

Step 1.3.3.3

Multiply $10$ by $1$ .

$2 y y' + 10$

Step 1.4

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

$3 x^{2} + 4 y^{3} y' = 2 y y' + 10$

Step 1.5

Solve for $y'$ .

Tap for more steps...

Step 1.5.1

Subtract $2 y y'$ from both sides of the equation.

$3 x^{2} + 4 y^{3} y' - 2 y y' = 10$

Step 1.5.2

Subtract $3 x^{2}$ from both sides of the equation.

$4 y^{3} y' - 2 y y' = 10 - 3 x^{2}$

Step 1.5.3

Factor $2 y y'$ out of $4 y^{3} y' - 2 y y'$ .

Tap for more steps...

Step 1.5.3.1

Factor $2 y y'$ out of $4 y^{3} y'$ .

$2 y y' (2 y^{2}) - 2 y y' = 10 - 3 x^{2}$

Step 1.5.3.2

Factor $2 y y'$ out of $- 2 y y'$ .

$2 y y' (2 y^{2}) + 2 y y' \cdot - 1 = 10 - 3 x^{2}$

Step 1.5.3.3

Factor $2 y y'$ out of $2 y y' (2 y^{2}) + 2 y y' \cdot - 1$ .

$2 y y' (2 y^{2} - 1) = 10 - 3 x^{2}$

Step 1.5.4

Divide each term in $2 y y' (2 y^{2} - 1) = 10 - 3 x^{2}$ by $2 y (2 y^{2} - 1)$ and simplify.

Tap for more steps...

Step 1.5.4.1

Divide each term in $2 y y' (2 y^{2} - 1) = 10 - 3 x^{2}$ by $2 y (2 y^{2} - 1)$ .

$\frac{2 y y' (2 y^{2} - 1)}{2 y (2 y^{2} - 1)} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2

Simplify the left side.

Tap for more steps...

Step 1.5.4.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.5.4.2.1.1

Cancel the common factor.

$\frac{2 y y' (2 y^{2} - 1)}{2 y (2 y^{2} - 1)} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2.1.2

Rewrite the expression.

$\frac{y y' (2 y^{2} - 1)}{y (2 y^{2} - 1)} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 1.5.4.2.2.1

Cancel the common factor.

$\frac{y y' (2 y^{2} - 1)}{y (2 y^{2} - 1)} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2.2.2

Rewrite the expression.

$\frac{y' (2 y^{2} - 1)}{2 y^{2} - 1} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2.3

Cancel the common factor of $2 y^{2} - 1$ .

Tap for more steps...

Step 1.5.4.2.3.1

Cancel the common factor.

$\frac{y' (2 y^{2} - 1)}{2 y^{2} - 1} = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{10}{2 y (2 y^{2} - 1)} + \frac{- 3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.5.4.3

Simplify the right side.

Tap for more steps...

Step 1.5.4.3.1

Simplify each term.

Tap for more steps...

Step 1.5.4.3.1.1

Cancel the common factor of $10$ and $2$ .

Tap for more steps...

Step 1.5.4.3.1.1.1

Factor $2$ out of $10$ .

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

Step 1.5.4.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 1.5.4.3.1.1.2.1

Factor $2$ out of $2 y (2 y^{2} - 1)$ .

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

Step 1.5.4.3.1.1.2.2

Cancel the common factor.

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

Step 1.5.4.3.1.1.2.3

Rewrite the expression.

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

Step 1.5.4.3.1.2

Move the negative in front of the fraction.

$y' = \frac{5}{y (2 y^{2} - 1)} - \frac{3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.6

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

$\frac{d y}{d x} = \frac{5}{y (2 y^{2} - 1)} - \frac{3 x^{2}}{2 y (2 y^{2} - 1)}$

Step 1.7

Evaluate at $x = 0$ and $y = 1$ .

Tap for more steps...

Step 1.7.1

Replace the variable $x$ with $0$ in the expression.

$\frac{5}{y (2 y^{2} - 1)} - \frac{3 {(0)}^{2}}{2 y (2 y^{2} - 1)}$

Step 1.7.2

Replace the variable $y$ with $1$ in the expression.

$\frac{5}{(1) (2 {(1)}^{2} - 1)} - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3

Simplify each term.

Tap for more steps...

Step 1.7.3.1

Multiply $2 \cdot 1^{2} - 1$ by $1$ .

$\frac{5}{2 \cdot 1^{2} - 1} - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3.2

Simplify the denominator.

Tap for more steps...

Step 1.7.3.2.1

One to any power is one.

$\frac{5}{2 \cdot 1 - 1} - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3.2.2

Multiply $2$ by $1$ .

$\frac{5}{2 - 1} - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3.2.3

Subtract $1$ from $2$ .

$\frac{5}{1} - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3.3

Divide $5$ by $1$ .

$5 - \frac{3 {(0)}^{2}}{2 (1) (2 {(1)}^{2} - 1)}$

Step 1.7.3.4

Raising $0$ to any positive power yields $0$ .

$5 - \frac{3 \cdot 0}{2 (1) (2 \cdot 1^{2} - 1)}$

Step 1.7.3.5

Multiply $2$ by $1$ .

$5 - \frac{3 \cdot 0}{2 (2 \cdot 1^{2} - 1)}$

Step 1.7.3.6

Multiply $3$ by $0$ .

$5 - \frac{0}{2 (2 \cdot 1^{2} - 1)}$

Step 1.7.3.7

Simplify the denominator.

Tap for more steps...

Step 1.7.3.7.1

One to any power is one.

$5 - \frac{0}{2 (2 \cdot 1 - 1)}$

Step 1.7.3.7.2

Multiply $2$ by $1$ .

$5 - \frac{0}{2 (2 - 1)}$

Step 1.7.3.7.3

Subtract $1$ from $2$ .

$5 - \frac{0}{2 \cdot 1}$

Step 1.7.3.8

Multiply $2$ by $1$ .

$5 - \frac{0}{2}$

Step 1.7.3.9

Divide $0$ by $2$ .

$5 - 0$

Step 1.7.3.10

Multiply $- 1$ by $0$ .

$5 + 0$

Step 1.7.4

Add $5$ and $0$ .

$5$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

Tap for more steps...

Step 2.1

Use the slope $5$ and a given point $(0, 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 5 \cdot (x - (0))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y - 1 = 5 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

Tap for more steps...

Step 2.3.1

Add $x$ and $0$ .

$y - 1 = 5 x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = 5 x + 1$

Step 3