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Secant of a Curve y =f(x)

Find the slop...

Question

Find the slope of the tangent and normal to the curve $y = (sin 2 x + cot x + 2)^{2}$ at $x = \frac{π}{2}$ .

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Solution

We have,
$y = (sin 2 x + cot x + 2)^{2}$
On differentiating both sides w.r.t. x, we get,
$\frac{d y}{d x} = 2 (sin 2 x + cot x + 2) \frac{d}{d x} (sin 2 x + cot x + 2)$

$\frac{d y}{d x} = 2 (sin 2 x + cot x + 2) (2 cos 2 x - c o s e c^{2} x)$

Therefore, $(\frac{d y}{d x})_{x = π / 2} = sin π + cot \frac{π}{2} + 2$

$= 2 [0 + 0 + 2] [- 2 - 1] = - 12$
Therefore, slope of tangent = -12
And, slope of normal $= \frac{1}{12}$

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