Find dy/dx y=(8x^3+9)^(3/2)

$y = {(8 x^{3} + 9)}^{\frac{3}{2}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(8 x^{3} + 9)}^{\frac{3}{2}})$

Step 2

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

$y'$

Step 3

Differentiate the right side of the equation.

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{3}{2}}$ and $g (x) = 8 x^{3} + 9$ .

To apply the Chain Rule, set $u$ as $8 x^{3} + 9$ .

$\frac{d}{d u} [u^{\frac{3}{2}}] \frac{d}{d x} [8 x^{3} + 9]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{3}{2}$ .

$\frac{3}{2} u^{\frac{3}{2} - 1} \frac{d}{d x} [8 x^{3} + 9]$

Replace all occurrences of $u$ with $8 x^{3} + 9$ .

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{3}{2} - 1} \frac{d}{d x} [8 x^{3} + 9]$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{3}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [8 x^{3} + 9]$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{3}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [8 x^{3} + 9]$

Combine the numerators over the common denominator.

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{3 - 1 \cdot 2}{2}} \frac{d}{d x} [8 x^{3} + 9]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{3 - 2}{2}} \frac{d}{d x} [8 x^{3} + 9]$

Subtract $2$ from $3$ .

$\frac{3}{2} {(8 x^{3} + 9)}^{\frac{1}{2}} \frac{d}{d x} [8 x^{3} + 9]$

Combine $\frac{3}{2}$ and ${(8 x^{3} + 9)}^{\frac{1}{2}}$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} \frac{d}{d x} [8 x^{3} + 9]$

By the Sum Rule, the derivative of $8 x^{3} + 9$ with respect to $x$ is $\frac{d}{d x} [8 x^{3}] + \frac{d}{d x} [9]$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (\frac{d}{d x} [8 x^{3}] + \frac{d}{d x} [9])$

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{3}$ with respect to $x$ is $8 \frac{d}{d x} [x^{3}]$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (8 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [9])$

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

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (8 (3 x^{2}) + \frac{d}{d x} [9])$

Multiply $3$ by $8$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (24 x^{2} + \frac{d}{d x} [9])$

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (24 x^{2} + 0)$

Simplify terms.

Add $24 x^{2}$ and $0$ .

$\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} (24 x^{2})$

Combine $24$ and $\frac{3 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2}$ .

$\frac{24 (3 {(8 x^{3} + 9)}^{\frac{1}{2}})}{2} x^{2}$

Multiply $3$ by $24$ .

$\frac{72 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2} x^{2}$

Combine $\frac{72 {(8 x^{3} + 9)}^{\frac{1}{2}}}{2}$ and $x^{2}$ .

$\frac{72 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}}{2}$

Factor $2$ out of $72 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}$ .

$\frac{2 (36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2})}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{2 (36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2})}{2 (1)}$

Cancel the common factor.

$\frac{2 (36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2})}{2 \cdot 1}$

Rewrite the expression.

$\frac{36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}}{1}$

Divide $36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}$ by $1$ .

$36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}$

Reorder the factors of $36 {(8 x^{3} + 9)}^{\frac{1}{2}} x^{2}$ .

$36 x^{2} {(8 x^{3} + 9)}^{\frac{1}{2}}$

Step 4

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

$y' = 36 x^{2} {(8 x^{3} + 9)}^{\frac{1}{2}}$

Step 5

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

$\frac{d y}{d x} = 36 x^{2} {(8 x^{3} + 9)}^{\frac{1}{2}}$

Find dy/dx y=(8x^3+9)^(3/2)

Related Questions

Support LCMGCF.com by disabling your adblocker.