Question

differentiate the following function.

$f(x)=\sqrt{e^{23x}-12}$

$\frac{d}{dx}(\sqrt{e^{23x}-12})=\square$

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\sqrt{e^{23x}-12}=(e^{23x} - 12)^{\frac{1}{2}}\) using the rule \( \sqrt{a}=a^{\frac{1}{2}}\).

Step2: Apply the chain rule

The chain rule states that if \( y = f(g(x))\), then \( y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \( u = e^{23x}-12\), so \( y = u^{\frac{1}{2}}\). First, find \( \frac{dy}{du}\):

$$ \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}} $$

Then, find \( \frac{du}{dx}\):
Since \( \frac{d}{dx}(e^{ax})=ae^{ax}\) (where \( a = 23\) here), \( \frac{du}{dx}=23e^{23x}\)

Step3: Calculate the derivative

By the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \( u = e^{23x}-12\), \( \frac{dy}{du}=\frac{1}{2\sqrt{e^{23x}-12}}\) and \( \frac{du}{dx}=23e^{23x}\)

$$ \frac{dy}{dx}=\frac{23e^{23x}}{2\sqrt{e^{23x}-12}} $$

Answer:

\(\frac{23e^{23x}}{2\sqrt{e^{23x}-12}}\)