If y-cos 2 x2- find d y-d x-

Y=(cos x^2)^2. put x^2=t and cos x^2=cos t=u, so that y=u^2, u=cos t and t=x^2 ∴dy/du=2u, du/dt= sin t and dt/dx=2x. So, dy/dx=(dy/du×du/dt×dt/dx) =4ux sin ...

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Byju's Answer

Standard XI

Mathematics

Algebra of Derivatives

If y=cos 2 x2...

Question

If y= $c o s^{2} x^{2}, f i n d \frac{d y}{d x} .$

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Solution

$y = (c o s x^{2})^{2} . p u t x^{2} = t a n d c o s x^{2} = c o s t = u,$ so that

$y = u^{2}, u = c o s t a n d t = x^{2}$

$∴ \frac{d y}{d u} = 2 u, \frac{d u}{d t} = - s i n t a n d \frac{d t}{d x} = 2 x .$

So, $\frac{d y}{d x} = (\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x})$

$= 4 u x s i n t = - 4 x s i n t c o s t [∵ u = c o s t]$

$= - 4 x s i n x^{2} c o s x^{2} = - 2 x s i n (2 x^{2}) [∵ t = x^{2}]$

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If y=cos2 x2,finddydx.

y=(cos x2)2.put x2=t and cos x2=cos t=u, so that y=u2,u=cos t and t=x2 ∴dydu=2u,dudt=−sin t and dtdx=2x. So, dydx=(dydu×dudt×dtdx) =4ux sin t=−4x sin t cos t [∵u=cost] =−4x sin x2cos x2=−2x sin(2x2) [∵t=x2]

If y= $c o s^{2} x^{2}, f i n d \frac{d y}{d x} .$