Find dy/dx y^4-x^3=7x

$y^{4} - x^{3} = 7 x$

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{4} - x^{3}) = \frac{d}{d x} (7 x)$

Differentiate the left side of the equation.

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

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

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

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 $g (x) = y$ .

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

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

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

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

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

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

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

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

Evaluate $\frac{d}{d x} [- x^{3}]$ .

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

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

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

$4 y^{3} y' - (3 x^{2})$

Multiply $3$ by $- 1$ .

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

Reorder terms.

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

Differentiate the right side of the equation.

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x]$

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

$7 \cdot 1$

Multiply $7$ by $1$ .

$7$

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

$- 3 x^{2} + 4 y^{3} y' = 7$

Solve for $y'$ .

Add $3 x^{2}$ to both sides of the equation.

$4 y^{3} y' = 7 + 3 x^{2}$

Divide each term in $4 y^{3} y' = 7 + 3 x^{2}$ by $4 y^{3}$ and simplify.

Divide each term in $4 y^{3} y' = 7 + 3 x^{2}$ by $4 y^{3}$ .

$\frac{4 y^{3} y'}{4 y^{3}} = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{3} y'}{4 y^{3}} = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

Rewrite the expression.

$\frac{y^{3} y'}{y^{3}} = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

Cancel the common factor of $y^{3}$ .

Cancel the common factor.

$\frac{y^{3} y'}{y^{3}} = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

Divide $y'$ by $1$ .

$y' = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

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

$\frac{d y}{d x} = \frac{7}{4 y^{3}} + \frac{3 x^{2}}{4 y^{3}}$

Find dy/dx y^4-x^3=7x

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