If y-tan-logx2- then dydx is -

Y=tan[log(x^2)] differentiate w.r.t x dydx=ddx[tan{log(x^2)}] dydx=sec^2(log x^2).ddx[log(x^2)] dydx=sec^2 (log x^2).1x^2.ddx(x^2) dydx=^2 (log x^2).1x^2.2x dyd ...

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Byju's Answer

Standard XIII

Physics

Chain Rule of Differentiation

If y=tan[logx...

Question

If $y = tan [log (x^{2})]$ , then $\frac{d y}{d x}$ is -

2 x {sec}^{2} [log (x^{2})]

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\frac{2 {sec}^{2} [log (x^{2})]}{x}

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x {sec}^{2} [log (x^{2})]

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\frac{1}{x^{2}} {sec}^{2} [log (x^{2})]

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Solution

The correct option is B $\frac{2 {sec}^{2} [log (x^{2})]}{x}$
$y = tan [log (x^{2})]$ differentiate w.r.t $x$
$\frac{d y}{d x} = \frac{d}{d x} [tan {log (x^{2})}]$
$\frac{d y}{d x} = s e c^{2} (log x^{2}) . \frac{d}{d x} [log (x^{2})]$
$\frac{d y}{d x} = s e c^{2} (log x^{2}) . \frac{1}{x^{2}} . \frac{d}{d x} (x^{2})$
$\frac{d y}{d x} = {sec}^{2} (log x^{2}) . \frac{1}{x^{2}} .2 x$
$\frac{d y}{d x} = \frac{2}{x} . s e c^{2} (log x^{2})$

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If y=tan[log(x2)], then dydx is -

The correct option is B 2 sec2[log(x2)]xy=tan[log(x2)] differentiate w.r.t x dydx=ddx[tan{log(x2)}] dydx=sec2(log x2).ddx[log(x2)] dydx=sec2(log x2).1x2.ddx(x2) dydx=sec2(log x2).1x2.2x dydx=2x.sec2(log x2)

If $y = tan [log (x^{2})]$ , then $\frac{d y}{d x}$ is -