Find the Derivative - d/dx y=e^(2x^(5-8))

$y = e^{2 x^{5 - 8}}$

Step 1

Subtract $8$ from $5$ .

$\frac{d}{d x} [e^{2 x^{- 3}}]$

Step 2

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

To apply the Chain Rule, set $u$ as $2 x^{- 3}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x^{- 3}]$

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

$e^{u} \frac{d}{d x} [2 x^{- 3}]$

Replace all occurrences of $u$ with $2 x^{- 3}$ .

$e^{2 x^{- 3}} \frac{d}{d x} [2 x^{- 3}]$

Step 3

Differentiate.

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{- 3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{- 3}]$ .

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

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

$e^{2 x^{- 3}} (2 (- 3 x^{- 4}))$

Multiply $- 3$ by $2$ .

$e^{2 x^{- 3}} (- 6 x^{- 4})$

Step 4

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$e^{2 \frac{1}{x^{3}}} (- 6 x^{- 4})$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$e^{2 \frac{1}{x^{3}}} (- 6 \frac{1}{x^{4}})$

Combine terms.

Combine $2$ and $\frac{1}{x^{3}}$ .

$e^{\frac{2}{x^{3}}} (- 6 \frac{1}{x^{4}})$

Combine $- 6$ and $\frac{1}{x^{4}}$ .

$e^{\frac{2}{x^{3}}} \frac{- 6}{x^{4}}$

Move the negative in front of the fraction.

$e^{\frac{2}{x^{3}}} (- \frac{6}{x^{4}})$

Combine $e^{\frac{2}{x^{3}}}$ and $\frac{6}{x^{4}}$ .

$- \frac{e^{\frac{2}{x^{3}}} \cdot 6}{x^{4}}$

Move $6$ to the left of $e^{\frac{2}{x^{3}}}$ .

$- \frac{6 e^{\frac{2}{x^{3}}}}{x^{4}}$

Find the Derivative - d/dx y=e^(2x^(5-8))

Related Questions

Support LCMGCF.com by disabling your adblocker.