\(\lim\limits_{x \rightarrow 0}\frac{sin^2?x}{x^2}\) =
= (\(\lim\limits_{x \rightarrow 0}\frac{sin ?x}{x}\) x \(\lim\limits_{x \rightarrow 0}\frac{sinx}{x}\))
We apply L?Hospital?s rule and differentiate numerator and denominator.
= (\(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\) x \(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\))
= 1

This is of form \(\frac{0}{0}\), refore we use L?Hospital?s rule and differentiate numerator and denominator.
= \(\lim\limits_{x \rightarrow 0}\frac{2sin? \,x cos \,?x + cos \,?x \sqrt 2}{2x ? 4}\)
= \(\frac{0+\sqrt 2}{-4}\)
= \(\frac{-1}{2\sqrt 2}\)

We need to use product rule in both terms to get answer.
\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f) + f.\(\frac{d}{dx}\) (g)
Here f = cos²? x and g = tan ?x
To differentiate f, we need to use chain rule.
\(\frac{d}{dx}\) (cos² ?x tan? x) = tan ?x.\(\frac{d}{dx}\) (cos² x) + cos² x.\(\frac{d}{dx}\) (tan? x)
\(\frac{d}{dx}\) (e^x tan?x) = tan? x.(-2 cos? x sin? x) + cos² ?x.sec² x
At x = 1 we get,
= tan?0.(-2 cos? 0 sin? 0) + cos²? 0.sec² ?0
= 1

We have, y = tan^-1(x/a)
Differentiating two times, with respect to x, we get,
y? = d/dx(tan^-1(x/a))
= 1/(1 + (x/a)²)*d/dx(x/a)
= a²/(x² + a²)* 1/a
= a/(x² + a²)
y? = d/dx (a/(x² + a²))
= a* d/dx(x² + a²)^-1
= a(-1) (x² + a²)^-2*d/dx(x² + a²)
Now solving we get,
= -a/(x² + a²)²*2x
= -(2ax/(x² + a²)²)

Clearly f (x) is not differentiable at x = 1 and x = -1
And x = 0 is not a critical point not in domain.
Therefore 1 and -1 are critical points.
Thus, re are 2 critical points.

The denominator becomes 0, as x approaches 4.
\(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\) Here, if we factorize numerator we get:
\(\lim\limits_{x \rightarrow 4} \frac{(x – 4)(x + 2)}{x – 4}\)
We can now cancel out (x ? 4) from both numerator and denominator.
We get, \(\lim\limits_{x \rightarrow 4}\)(x + 2) = 6

Clearly, x = 1, 3 are points of global minimum (as values being equal).
And, x = 0, 4 are points of global maximum (as values being equal).
And, x = 2, is point of local maximum (as values being equal).
Thus, x = 3 is global maximum.

\(\lim\limits_{x \rightarrow 0}(\frac{1}{e^x} + sin? x) = \frac{1}{e^0}\) + cos (0)
= 1 + 1
= 2

We have, y = tan^-1 √(x² ? 1)
Differentiating both sides with respect to x, we get,
y? = 1/(1 + x² ? 1)* d/dx(x² ? 1)^1/2
1/x² * 1/2((x² ? 1))^1/2 * 2x
Squaring both sides,
x²(x² ? 1)(y?)² = 1
Differentiating again with respect to x we get,
x²(x² ? 1)(y?)² + (y?)²d/dx(x⁴ ? x²) = d/dx(1)
Or, x²(x² ? 1)²y?y? + (y?)²(4x³ ? 2x) = 0
Or, x²(x² ? 1)y? + (y?)(2x³ ? x) = 0

Using quotient rule, we know that, \(\frac{d}{dx} (\frac{f}{g}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)
Here, f = a^x and g = e^x
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x.\frac{d}{dx}(a^x)-a^x.\frac{d}{dx}(e^x)}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x a^x ln?\, a-a^x e^x}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{a^x (ln? \,a-1)}{e^x}\)