a. Find from first principles, the derivative of (bx a)^n b. Find the area of the circle x^2 + y^2 = 36 , using method of integration.

a. Derivative of $(bx-a{)}^n$ from First Principles

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

Let $f(x)=(bx-a{)}^n$. Then, $$f(x+h)=[b(x+h)-a{]}^n=(bx+bh-a{)}^n$$

Now, compute the difference quotient: $$\frac{f(x+h)-f(x)}{h}=\frac{(bx+bh-a{)}^n-(bx-a{)}^n}{h}$$

Factor out $(bx-a{)}^n$: $$=(bx-a{)}^n\cdot \frac{{\left(1+\frac{bh}{bx-a}\right)}^n-1}{h}$$

Let $k=\frac{bh}{bx-a}$, so $h=\frac{k(bx-a)}{b}$. Substituting: $$=(bx-a{)}^n\cdot \frac{(1+k{)}^n-1}{\frac{k(bx-a)}{b}}=\frac{b(bx-a{)}^{n-1}\left[(1+k{)}^n-1\right]}{k}$$

Using the binomial expansion for $(1+k{)}^n$ when $k\to 0$ (i.e., $h\to 0$): $$(1+k{)}^n\approx 1+nk+\frac{n(n-1)}{2}{k}^2+\cdots$$ Thus, $$(1+k{)}^n-1\approx nk+\frac{n(n-1)}{2}{k}^2$$

Substituting back: $$\frac{b(bx-a{)}^{n-1}\left[nk+\frac{n(n-1)}{2}{k}^2\right]}{k}=b(bx-a{)}^{n-1}\left[n+\frac{n(n-1)}{2}k\right]$$

Taking the limit as $h\to 0$ (or $k\to 0$): $$f^{\prime }(x)=\underset{k\to 0}{\lim }b(bx-a{)}^{n-1}\left[n+\frac{n(n-1)}{2}k\right]=bn(bx-a{)}^{n-1}$$

Final answer: $$\boxed{bn(bx-a{)}^{n-1}}$$

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b. Area of the Circle $x^2+y^2=36$ Using Integration

The equation $x^2+y^2=36$ represents a circle centered at the origin with radius $r=6$.

Step 1: Express $y$ in terms of $x$

$$y=\pm \sqrt{36-x^2}$$

Step 2: Find the limits of integration

The circle intersects the x-axis at $x=\pm 6$. Thus, the limits are from $-6$ to $6$.

Step 3: Area using integration

The area $A$ of the circle is twice the area above the x-axis (due to symmetry): $$A=2{\int }_{-6}^6\sqrt{36-x^2}dx$$

Step 4: Use trigonometric substitution

Let $x=6\sin \theta$, then $dx=6\cos \theta d\theta$. When $x=-6$, $\theta =-\frac{\pi }{2}$; when $x=6$, $\theta =\frac{\pi }{2}$.

Substitute: $$\sqrt{36-x^2}=\sqrt{36-36{\sin }^2\theta }=6\cos \theta$$

Thus, $$A=2{\int }_{-\pi /2}^{\pi /2}6\cos \theta \cdot 6\cos \theta d\theta =72{\int }_{-\pi /2}^{\pi /2}{\cos }^2\theta d\theta$$

Using the identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$: $$A=72{\int }_{-\pi /2}^{\pi /2}\frac{1+\cos 2\theta }{2}d\theta =36{\left[\theta +\frac{\sin 2\theta }{2}\right]}_{-\pi /2}^{\pi /2}$$

Evaluate the integral: $$A=36\left[\left(\frac{\pi }{2}+0\right)-\left(-\frac{\pi }{2}+0\right)\right]=36\cdot \pi =36\pi$$

Final answer: $$\boxed{36\pi }$$