Y-sin-x-cos 2-xFind d y-d x

Y=sin√(x)+(cos√(x))^2 ∴d/dx [sin ( x^1/2 ) ]+d/dx [cos ( x^1/2 ) ] =cos ( x^1/2 ).d/dxx^1/2+2cos ( x^1/2 )d/dx [ cos ( x^1/2 ) ] =cos ( x^1/2 )1/2x^ 1/2+ ...

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Byju's Answer

Standard XII

Mathematics

Inequality

y=sin√x+cos 2...

Question

$y = s i n \sqrt{x} + c o s^{2} \sqrt{x}$

Find $\frac{d y}{d x}$

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Solution

$y = s i n \sqrt{x} + (c o s \sqrt{x})^{2} ∴ \frac{d}{d x} [s i n (x^{\frac{1}{2}})] + \frac{d}{d x} [c o s (x^{\frac{1}{2}})] = c o s (x^{\frac{1}{2}}) . \frac{d}{d x} x^{\frac{1}{2}} + 2 c o s (x^{\frac{1}{2}}) \frac{d}{d x} [c o s (x^{\frac{1}{2}})] = c o s (x^{\frac{1}{2}}) \frac{1}{2} x^{\frac{- 1}{2}} + 2. c o s (x^{\frac{1}{2}}) . [- s i n (x^{\frac{1}{2}} . \frac{d}{d x} x^{\frac{1}{2}})] = c o s \sqrt{x} . \frac{1}{2 \sqrt{x}} [- 2 c o s (x^{\frac{1}{2}})] . s i n x^{\frac{1}{2}} . \frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{x}} [c o s (\sqrt{x}) - s i n (2 \sqrt{x})]$

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Similar questions

Q. Find

\frac{d y}{d x}

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x = cos θ - cos 2 θ

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{sin}^{2} x + {cos}^{2} y = 1

, then

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is equal to

Q. Solve:

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y = sin sin x \frac{d y}{d x} = ?

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(tan y) \frac{d y}{d x} = sin (x + y) + sin (x - y)

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y=sin√x+cos2√x Find dydx

y=sin√x+(cos√x)2∴ddx[sin(x12)]+ddx[cos(x12)]=cos(x12).ddxx12+2cos(x12)ddx[cos(x12)]=cos(x12)12x−12+2.cos(x12).[−sin(x12.ddxx12)]=cos√x.12√x[−2cos(x12)].sinx12.12√x=12√x[cos(√x)−sin(2√x)]

$y = s i n \sqrt{x} + c o s^{2} \sqrt{x}$

Find $\frac{d y}{d x}$