We know that in range of (0, 2π) graph of sinx and cosx intersects each or in three points.
And we know that se points of intersection are only critical points
Thus, re are 3 critical points.

Let A, B, C be vertices of isosceles triangle.
Let B = (a cos&ta;, b sin&ta;)
=> C – (a cos&ta;. -b sin&ta;)  (by symmetry).
Now, area of a triangle is given by
A = �(BC)(AD)
= �(2b sin&ta;)(1 + cos&ta;)
=> dA/d&ta; = ab[cos&ta;(1 + cos&ta;) ? sin²&ta;]
= ab(1 + cos&ta;)(2 cos&ta; ? 1)
Now, putting dA/d&ta; = 0,
We get, &ta; = π/3, π
Now, for &ta; = π, triangle ABC is not possible.
And, dA/d&ta;]_{&ta; = π/3} < 0,
Thus, for &ta; = π/3, area of isosceles triangle will be maximum.
So, A = 3√3/4 ab sq units.

Since it is of form \(\frac{\infty}{\infty}\), we use L?Hospital?s rule and differentiate numerator and denominator
L = \(\lim\limits_{x \rightarrow \infty}\frac{x^2-9}{x^2?3x+2}\)
On differentiating once, we get \(\lim\limits_{x \rightarrow \infty}\frac{2x}{2x}\)
Which is equal to, \(\lim\limits_{x \rightarrow \infty}\) ?1 = 1.

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

We need to use product rule in both terms to get answer.
\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)
\(\frac{d}{dx}\) (e^x sin x + e^x cos ?x) = (e^x.\(\frac{d}{dx}\) (sin? x) + sin ?x.\(\frac{d}{dx}\) (e^x)) + (e^x.\(\frac{d}{dx}\) (cos ?x) + cos? x.\(\frac{d}{dx}\) (e^x))
\(\frac{d}{dx}\) (e^x sin x + e^x cos ?x) =(e^x.cos? x + sin ?x . e^x) + (e^x.(-sin ?x) + cos ?x.e^x)
\(\frac{d}{dx}\) (e^x sin x + e^x cos ?x) = e^x.cos? x + sin? x . e^x – e^x.sin? x + cos ?x.e^x
\(\frac{d}{dx}\) (e^x sin x + e^x cos ?x) = 2e^x.cos? x

\(\frac{d}{dx}(\frac{sec?x}{cosec\, x\, tan?x})\) = \(\frac{d}{dx}(\frac{{1} {cos?x}}{\frac{1}{sin?\, x}.\frac{sin?x}{cos?x}}\))
= \(\frac{d}{dx}\) (1)
= 0

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

We have, y = cos(2sin^-1x)
Differentiating both sides,
y? = d/dxcos(2sin^-1x)
= -sin(2sin^-1x) * d/dx(2sin^-1x)
or, y? = -sin(2sin^-1x)*2*1/√(1 ? x²)
or, y?*√(1 ? x²) = -2sin(2sin^-1x)
Squaring both sides,
or, (y?)² *(1 ? x²) = 4sin²(2sin^-1x)
or, (y?)² *(1 ? x²) = 4(1 ? y²)
Differentiating again with respect to x, we get,
(1 ? x²)*2y?y? + (y?)² (-2x) = 4 * (-2yy?)
(1 ? x²)y? ? xy? = – 4y
=> (1 ? x²)y? ? xy? + 4y = 0

Given, y = 1/(1 + x + x² + x³)
Assuming x ≠ 1
Or, y = (x ? 1)/(x ? 1)( x³ + x² + x + 1)
Differentiating both sides with respect to x, we get,
dy/dx = [(x⁴ ? 1)*1 ? (x ? 1)*4x³]/ (x⁴ ? 1)²
Thus, putting x = 0 in above equation, we get,
(dy/dx) = -1/(-1)²
= -1

This is of form \(\frac{\infty}{\infty}\), refore we use L?Hospital?s rule and differentiate numerator and denominator.
= \(\lim\limits_{x \rightarrow \infty}\frac{2x+\sqrt{2/x}}{2x?4}\)
= \(\lim\limits_{x \rightarrow \infty}?\)√2 x^-3/2
= 0