find dy for the given values of x and δx. y = √(9x + 2); x = 5; δx = 0.25 what is the derivative of y = √(9x…

Explanation:

Step1: Use chain - rule for derivative

Let $u = 9x + 2$, then $y=\sqrt{u}=u^{\frac{1}{2}}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$, and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=9$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $\frac{dy}{dx}=\frac{9}{2\sqrt{9x + 2}}$.

Step2: Calculate $dy$

We know that $dy=\frac{dy}{dx}\Delta x$. Substitute $x = 5$ into $\frac{dy}{dx}$, we get $\frac{dy}{dx}\big|{x = 5}=\frac{9}{2\sqrt{9\times5+2}}=\frac{9}{2\sqrt{47}}$. Then substitute $\frac{dy}{dx}\big|{x = 5}$ and $\Delta x=0.25$ into $dy=\frac{dy}{dx}\Delta x$. So $dy=\frac{9}{2\sqrt{47}}\times0.25\approx\frac{9}{2\times6.856}\times0.25=\frac{9}{13.712}\times0.25\approx0.164$.

Answer:

$\frac{dy}{dx}=\frac{9}{2\sqrt{9x + 2}}$; $dy\approx0.164$