Questions: Given (y=f(u)) and (u=g(x)), find (fracd yd x) by using Leibniz's notation for the chain rule, (fracd yd x=fracd yd u fracd ud x). (y=sqrt2 u+4, u=x^2-9 x) (fracd yd x=square)

$Given (y=f(u)) and (u=g(x)), find (fracd yd x) by using Leibniz's notation for the chain rule, (fracd yd x=fracd yd u fracd ud x). (y=sqrt2 u+4, u=x^2-9 x) (fracd yd x=square)$

Transcript text: Given $y=f(u)$ and $u=g(x)$, find $\frac{d y}{d x}$ by using Leibniz's notation for the chain rule, $\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}$. \[ \begin{array}{c} y=\sqrt{2 u+4}, u=x^{2}-9 x \\ \frac{d y}{d x}=\square \end{array} \]

Solution

Solution Steps

To find $\frac{d y}{d x}$, we will use the chain rule in Leibniz's notation. First, we need to find $\frac{d y}{d u}$ by differentiating $y = \sqrt{2u + 4}$ with respect to $u$. Then, we find $\frac{d u}{d x}$ by differentiating $u = x^2 - 9x$ with respect to $x$. Finally, we multiply these derivatives together to get $\frac{d y}{d x}$.

Step 1: Differentiate $ y $ with respect to $ u $

Given $ y = \sqrt{2u + 4} $, we differentiate with respect to $ u $:

\[ \frac{d y}{d u} = \frac{1}{2} \cdot (2u + 4)^{-\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2u + 4}} \]

Step 2: Differentiate $ u $ with respect to $ x $

Given $ u = x^2 - 9x $, we differentiate with respect to $ x $:

\[ \frac{d u}{d x} = 2x - 9 \]

Step 3: Apply the Chain Rule

Using the chain rule $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$, we substitute the derivatives:

\[ \frac{d y}{d x} = \frac{1}{\sqrt{2u + 4}} \cdot (2x - 9) = \frac{2x - 9}{\sqrt{2u + 4}} \]