Question

Find the slope of the tangent line at the point (0, 0) to the curve with equation e^xy+sinx2=cos⁡x+y

Answer 1

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We have,

$y = (\sin2x + \cot \: x + 2x)^{2}$

On differentiating both sides w.r.t. x, we get ,

$\frac{dy}{dx} = 2( \sin2x + \cot \: x + 2) \frac{d}{dx} ( \sin2x + \cot \: x + 2)$

$\frac{dy}{dx} = 2( \sin2x + \cot \: x + 2)(2 \cos2x - \cos\sec^{2} x)$

Therefore,

$( \frac{dy}{dx} ) x = \pi > 2 = sin \: \pi + \cot \frac{\pi}{2} + 2$

(here after pi we need to write / not > )

$= 2(0 + 0 + 2)( - 2 - 1) = - 12$

Therefore, slope of tangent = -12

And, slope of normal =

$\frac{1}{12}$

hope it helps uh!!