Find the Second Derivative f(x)=(1.25^(x/4))^16

$f (x) = {({1.25}^{\frac{x}{4}})}^{16}$

Find the first derivative.

Multiply the exponents in ${({1.25}^{\frac{x}{4}})}^{16}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f' (x) = \frac{d}{d x} ({1.25}^{\frac{x}{4} \cdot 16})$

Cancel the common factor of $4$ .

Factor $4$ out of $16$ .

$f' (x) = \frac{d}{d x} ({1.25}^{\frac{x}{4} \cdot (4 (4))})$

Cancel the common factor.

$f' (x) = \frac{d}{d x} ({1.25}^{\frac{x}{4} \cdot (4 \cdot 4)})$

Rewrite the expression.

$f' (x) = \frac{d}{d x} ({1.25}^{x \cdot 4})$

Move $4$ to the left of $x$ .

$f' (x) = \frac{d}{d x} ({1.25}^{4 x})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = {1.25}^{x}$ and $g (x) = 4 x$ .

To apply the Chain Rule, set $u$ as $4 x$ .

$f' (x) = \frac{d}{d u} ({1.25}^{u}) \frac{d}{d x} (4 x)$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $1.25$ .

$f' (x) = {1.25}^{u} \ln (1.25) \frac{d}{d x} (4 x)$

Replace all occurrences of $u$ with $4 x$ .

$f' (x) = {1.25}^{4 x} \ln (1.25) \frac{d}{d x} (4 x)$

Differentiate.

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f' (x) = {1.25}^{4 x} \ln (1.25) (4 \frac{d}{d x} (x))$

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

$f' (x) = {1.25}^{4 x} \ln (1.25) (4 \cdot 1)$

Simplify the expression.

Multiply $4$ by $1$ .

$f' (x) = {1.25}^{4 x} \ln (1.25) \cdot 4$

Move $4$ to the left of ${1.25}^{4 x} \ln (1.25)$ .

$f' (x) = 4 \cdot ({1.25}^{4 x} \ln (1.25))$

Find the second derivative.

Since $4 \ln (1.25)$ is constant with respect to $x$ , the derivative of $4 \cdot {1.25}^{4 x} \ln (1.25)$ with respect to $x$ is $4 \ln (1.25) \frac{d}{d x} [{1.25}^{4 x}]$ .

$f'' (x) = 4 \ln (1.25) \frac{d}{d x} ({1.25}^{4 x})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = {1.25}^{x}$ and $g (x) = 4 x$ .

To apply the Chain Rule, set $u$ as $4 x$ .

$f'' (x) = 4 \ln (1.25) (\frac{d}{d u} ({1.25}^{u}) \frac{d}{d x} (4 x))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $1.25$ .

$f'' (x) = 4 \ln (1.25) ({1.25}^{u} \ln (1.25) \frac{d}{d x} (4 x))$

Replace all occurrences of $u$ with $4 x$ .

$f'' (x) = 4 \ln (1.25) ({1.25}^{4 x} \ln (1.25) \frac{d}{d x} (4 x))$

Raise $\ln (1.25)$ to the power of $1$ .

$f'' (x) = 4 ({1.25}^{4 x} ((\ln (1.25)) \ln (1.25)) \frac{d}{d x} (4 x))$

Raise $\ln (1.25)$ to the power of $1$ .

$f'' (x) = 4 ({1.25}^{4 x} ((\ln (1.25)) (\ln (1.25))) \frac{d}{d x} (4 x))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 4 ({1.25}^{4 x} {(\ln (1.25))}^{1 + 1} \frac{d}{d x} (4 x))$

Add $1$ and $1$ .

$f'' (x) = 4 ({1.25}^{4 x} \ln^{2} (1.25) \frac{d}{d x} (4 x))$

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = 4 \cdot ({1.25}^{4 x} \ln^{2} (1.25) (4 \frac{d}{d x} (x)))$

Multiply $4$ by $4$ .

$f'' (x) = 16 \cdot ({1.25}^{4 x} \ln^{2} (1.25)) \frac{d}{d x} (x)$

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

$f'' (x) = 16 \cdot ({1.25}^{4 x} \ln^{2} (1.25)) \cdot 1$

Multiply $16$ by $1$ .

$f'' (x) = 16 \cdot ({1.25}^{4 x} \ln^{2} (1.25))$

The second derivative of $f (x)$ with respect to $x$ is $16 \cdot ({1.25}^{4 x} \ln^{2} (1.25))$ .

$16 \cdot ({1.25}^{4 x} \ln^{2} (1.25))$

Find the Second Derivative f(x)=(1.25^(x/4))^16

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