Question

compute the derivative of the function ( y = sqrt{x^7 - 40x} ) using the chain rule. ( \frac{dy}{dx} = )

Explanation:

Step1: Rewrite the function

First, rewrite \( y = \sqrt{x^7 - 40x} \) as \( y=(x^7 - 40x)^{\frac{1}{2}} \). Let \( u = x^7 - 40x \), so \( y = u^{\frac{1}{2}} \).

Step2: Apply the Chain Rule

The Chain Rule states that \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \).

• Find \( \frac{dy}{du} \): Differentiate \( y = u^{\frac{1}{2}} \) with respect to \( u \). Using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \), we get \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}} \).
• Find \( \frac{du}{dx} \): Differentiate \( u = x^7 - 40x \) with respect to \( x \). Using the power rule, \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we have \( \frac{du}{dx}=7x^6 - 40 \).

Step3: Multiply the derivatives

Now, substitute \( u = x^7 - 40x \) back into \( \frac{dy}{du} \) and multiply by \( \frac{du}{dx} \):
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{1}{2}(x^7 - 40x)^{-\frac{1}{2}}\cdot(7x^6 - 40)\\ &=\frac{7x^6 - 40}{2\sqrt{x^7 - 40x}} \end{align*}$$

\]

Answer:

\(\frac{7x^{6}-40}{2\sqrt{x^{7}-40x}}\)