a) Find from first principles, the derivative of 2 3x . b) Find the area of the region between the curve y^2 = 4x and the line x = y .

a) Find from first principles, the derivative of $\sqrt{2-3x}$

Solution:

The derivative of a function $f(x)$ from first principles is defined as: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Let $f(x)=\sqrt{2-3x}$. We need to find $f^{\prime }(x)$.

Step 1: Write the difference quotient

$$f(x+h)=\sqrt{2-3(x+h)}=\sqrt{2-3x-3h}$$

The difference quotient is: $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}$$

Step 2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}\cdot \frac{\sqrt{2-3x-3h}+\sqrt{2-3x}}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$

This simplifies to: $$\frac{(2-3x-3h)-(2-3x)}{h\left(\sqrt{2-3x-3h}+\sqrt{2-3x}\right)}=\frac{-3h}{h\left(\sqrt{2-3x-3h}+\sqrt{2-3x}\right)}$$

Simplify the fraction: $$\frac{-3}{\sqrt{2-3x-3h}+\sqrt{2-3x}}$$

Step 3: Take the limit as $h\to 0$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-3}{\sqrt{2-3x-3h}+\sqrt{2-3x}}=\frac{-3}{2\sqrt{2-3x}}$$

Thus, the derivative of $\sqrt{2-3x}$ from first principles is: $$\boxed{\frac{d}{dx}\sqrt{2-3x}=-\frac{3}{2\sqrt{2-3x}}}$$

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b) Find the area of the region between the curve $y^2=4x$ and the line $x=y$.

Solution:

We need to find the area enclosed by the parabola $y^2=4x$ and the line $x=y$.

Step 1: Find the points of intersection

Set $x=y$ in the equation of the parabola: $$y^2=4y⟹y^2-4y=0⟹y(y-4)=0$$ Thus, $y=0$ or $y=4$.

The corresponding $x$-values are:

• For $y=0$, $x=0$.
• For $y=4$, $x=4$.

So, the points of intersection are $(0,0)$ and $(4,4)$.

Step 2: Express $x$ in terms of $y$ for both curves

• For the parabola $y^2=4x$, we have $x=\frac{y^2}{4}$.
• For the line $x=y$, we have $x=y$.

Step 3: Set up the integral for the area

The area $A$ between the curves from $y=0$ to $y=4$ is given by: $$A={\int }_0^4\left(\text{right function}-\text{left function}\right)dy$$ Here, the right function is $x=y$ and the left function is $x=\frac{y^2}{4}$.

Thus: $$A={\int }_0^4\left(y-\frac{y^2}{4}\right)dy$$

Step 4: Compute the integral

$$A={\int }_0^{4}ydy-\frac{1}{4}{\int }_0^4{y}^{2}dy$$

Compute each integral separately: $$\int ydy=\frac{y^2}{2},\int y^{2}dy=\frac{y^3}{3}$$

Evaluate from $0$ to $4$: $$A={\left[\frac{y^2}{2}\right]}_0^4-\frac{1}{4}{\left[\frac{y^3}{3}\right]}_0^4$$

Substitute the limits: $$A=\left(\frac{4^2}{2}-0\right)-\frac{1}{4}\left(\frac{4^3}{3}-0\right)$$ $$A=\frac{16}{2}-\frac{1}{4}\cdot \frac{64}{3}$$ $$A=8-\frac{16}{3}$$ $$A=\frac{24}{3}-\frac{16}{3}=\frac{8}{3}$$

Thus, the area of the region is: $$\boxed{\frac{8}{3}\text{ square units}}$$