Find the Second Derivative 5e^(-x)-6e^(-5x)

$5 e^{- x} - 6 e^{- 5 x}$

Step 1

Find the first derivative.

By the Sum Rule, the derivative of $5 e^{- x} - 6 e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [5 e^{- x}] + \frac{d}{d x} [- 6 e^{- 5 x}]$ .

$f' (x) = \frac{d}{d x} (5 e^{- x}) + \frac{d}{d x} (- 6 e^{- 5 x})$

Evaluate $\frac{d}{d x} [5 e^{- x}]$ .

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- x}]$ .

$f' (x) = 5 \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (- 6 e^{- 5 x})$

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 $g (x) = - x$ .

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$f' (x) = 5 (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x)) + \frac{d}{d x} (- 6 e^{- 5 x})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f' (x) = 5 (e^{u_{1}} \frac{d}{d x} (- x)) + \frac{d}{d x} (- 6 e^{- 5 x})$

Replace all occurrences of $u_{1}$ with $- x$ .

$f' (x) = 5 (e^{- x} \frac{d}{d x} (- x)) + \frac{d}{d x} (- 6 e^{- 5 x})$

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

$f' (x) = 5 (e^{- x} (- \frac{d}{d x} x)) + \frac{d}{d x} (- 6 e^{- 5 x})$

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

$f' (x) = 5 (e^{- x} (- 1 \cdot 1)) + \frac{d}{d x} (- 6 e^{- 5 x})$

Multiply $- 1$ by $1$ .

$f' (x) = 5 (e^{- x} \cdot - 1) + \frac{d}{d x} (- 6 e^{- 5 x})$

Move $- 1$ to the left of $e^{- x}$ .

$f' (x) = 5 (- 1 \cdot e^{- x}) + \frac{d}{d x} (- 6 e^{- 5 x})$

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$f' (x) = 5 (- e^{- x}) + \frac{d}{d x} (- 6 e^{- 5 x})$

Multiply $- 1$ by $5$ .

$f' (x) = - 5 e^{- x} + \frac{d}{d x} (- 6 e^{- 5 x})$

Evaluate $\frac{d}{d x} [- 6 e^{- 5 x}]$ .

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 e^{- 5 x}$ with respect to $x$ is $- 6 \frac{d}{d x} [e^{- 5 x}]$ .

$f' (x) = - 5 e^{- x} - 6 \frac{d}{d x} e^{- 5 x}$

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 $g (x) = - 5 x$ .

To apply the Chain Rule, set $u_{2}$ as $- 5 x$ .

$f' (x) = - 5 e^{- x} - 6 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- 5 x))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f' (x) = - 5 e^{- x} - 6 (e^{u_{2}} \frac{d}{d x} (- 5 x))$

Replace all occurrences of $u_{2}$ with $- 5 x$ .

$f' (x) = - 5 e^{- x} - 6 (e^{- 5 x} \frac{d}{d x} (- 5 x))$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$f' (x) = - 5 e^{- x} - 6 (e^{- 5 x} (- 5 \frac{d}{d x} x))$

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

$f' (x) = - 5 e^{- x} - 6 (e^{- 5 x} (- 5 \cdot 1))$

Multiply $- 5$ by $1$ .

$f' (x) = - 5 e^{- x} - 6 (e^{- 5 x} \cdot - 5)$

Move $- 5$ to the left of $e^{- 5 x}$ .

$f' (x) = - 5 e^{- x} - 6 (- 5 \cdot e^{- 5 x})$

Multiply $- 5$ by $- 6$ .

$f' (x) = - 5 e^{- x} + 30 e^{- 5 x}$

Reorder terms.

$f' (x) = 30 e^{- 5 x} - 5 e^{- x}$

Step 2

Find the second derivative.

By the Sum Rule, the derivative of $30 e^{- 5 x} - 5 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [30 e^{- 5 x}] + \frac{d}{d x} [- 5 e^{- x}]$ .

$f'' (x) = \frac{d}{d x} (30 e^{- 5 x}) + \frac{d}{d x} (- 5 e^{- x})$

Evaluate $\frac{d}{d x} [30 e^{- 5 x}]$ .

Since $30$ is constant with respect to $x$ , the derivative of $30 e^{- 5 x}$ with respect to $x$ is $30 \frac{d}{d x} [e^{- 5 x}]$ .

$f'' (x) = 30 \frac{d}{d x} (e^{- 5 x}) + \frac{d}{d x} (- 5 e^{- x})$

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 $g (x) = - 5 x$ .

To apply the Chain Rule, set $u_{1}$ as $- 5 x$ .

$f'' (x) = 30 (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- 5 x)) + \frac{d}{d x} (- 5 e^{- x})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = 30 (e^{u_{1}} \frac{d}{d x} (- 5 x)) + \frac{d}{d x} (- 5 e^{- x})$

Replace all occurrences of $u_{1}$ with $- 5 x$ .

$f'' (x) = 30 (e^{- 5 x} \frac{d}{d x} (- 5 x)) + \frac{d}{d x} (- 5 e^{- x})$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$f'' (x) = 30 (e^{- 5 x} (- 5 \frac{d}{d x} x)) + \frac{d}{d x} (- 5 e^{- x})$

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

$f'' (x) = 30 (e^{- 5 x} (- 5 \cdot 1)) + \frac{d}{d x} (- 5 e^{- x})$

Multiply $- 5$ by $1$ .

$f'' (x) = 30 (e^{- 5 x} \cdot - 5) + \frac{d}{d x} (- 5 e^{- x})$

Move $- 5$ to the left of $e^{- 5 x}$ .

$f'' (x) = 30 (- 5 \cdot e^{- 5 x}) + \frac{d}{d x} (- 5 e^{- x})$

Multiply $- 5$ by $30$ .

$f'' (x) = - 150 e^{- 5 x} + \frac{d}{d x} (- 5 e^{- x})$

Evaluate $\frac{d}{d x} [- 5 e^{- x}]$ .

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 e^{- x}$ with respect to $x$ is $- 5 \frac{d}{d x} [e^{- x}]$ .

$f'' (x) = - 150 e^{- 5 x} - 5 \frac{d}{d x} e^{- x}$

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 $g (x) = - x$ .

To apply the Chain Rule, set $u_{2}$ as $- x$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (e^{u_{2}} \frac{d}{d x} (- x))$

Replace all occurrences of $u_{2}$ with $- x$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (e^{- x} \frac{d}{d x} (- x))$

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

$f'' (x) = - 150 e^{- 5 x} - 5 (e^{- x} (- \frac{d}{d x} x))$

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

$f'' (x) = - 150 e^{- 5 x} - 5 (e^{- x} (- 1 \cdot 1))$

Multiply $- 1$ by $1$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (e^{- x} \cdot - 1)$

Move $- 1$ to the left of $e^{- x}$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (- 1 \cdot e^{- x})$

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$f'' (x) = - 150 e^{- 5 x} - 5 (- e^{- x})$

Multiply $- 1$ by $- 5$ .

$f'' (x) = - 150 e^{- 5 x} + 5 e^{- x}$

Reorder terms.

$f'' (x) = 5 e^{- x} - 150 e^{- 5 x}$

Find the Second Derivative 5e^(-x)-6e^(-5x)

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