Find the Derivative - d/d@VAR f(x)=(1.25^(x/4))^16

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

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

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

$\frac{d}{d x} [{1.25}^{\frac{x}{4} \cdot 16}]$

Cancel the common factor of $4$ .

Factor $4$ out of $16$ .

$\frac{d}{d x} [{1.25}^{\frac{x}{4} \cdot (4 (4))}]$

Cancel the common factor.

$\frac{d}{d x} [{1.25}^{\frac{x}{4} \cdot (4 \cdot 4)}]$

Rewrite the expression.

$\frac{d}{d x} [{1.25}^{x \cdot 4}]$

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

$\frac{d}{d x} [{1.25}^{4 x}]$

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 $4 x$ .

$\frac{d}{d u} [{1.25}^{u}] \frac{d}{d x} [4 x]$

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

${1.25}^{u} \ln (1.25) \frac{d}{d x} [4 x]$

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

${1.25}^{4 x} \ln (1.25) \frac{d}{d x} [4 x]$

Differentiate.

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

${1.25}^{4 x} \ln (1.25) (4 \frac{d}{d x} [x])$

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

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

Simplify the expression.

Multiply $4$ by $1$ .

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

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

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

Find the Derivative - d/d@VAR f(x)=(1.25^(x/4))^16

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