Question

if y=log(secx) findy3

Answer 1

y = log(secx)

differentiate with respect to x

dy/dx = d/dx(log(secx)
dy/dx = 1/secx × secx × tanx
dy/dx = tanx

Again differentiate with respect to x


y² = d/dx(tanx)
y² = sec²x

Again differentiate with respect to x


y³ = 2secx × secx.tanx ×1
y³ = 2 sec²x. tanx
and here it is your answer

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Answer 2

Answer:

Required value of $d {y}^{3}$

is $2 \tan(x) {sec}^{2} x$

Step-by-step explanation:

Given y= $log( \sec(x) )$

We are differentiating with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} ( log( \sec(x) ) = \frac{1}{secx} \times \frac{d}{dx} (secx)$

$= \frac{1 }{ \sec(x) } \times \sec(x) \tan(x)$

$= \tan(x)$

Again differentiating with respect to x and get,

$\frac{ {d}^{2}y }{d {y}^{2} } = \frac{d}{dx} (tanx) = {sec}^{2} x$

Again differentiating with respect to x and get,

$\frac{ {d}^{3}y }{d {x}^{3} } = \frac{d}{dx} ( {sec}^{2}x ) = 2secx \times \sec(x) \tan(x)$

$= 2 \tan(x) {sec}^{2} x$

Required value of $d {y}^{3} = 2 \tan(x) {sec}^{2} x$

Here applied formulas are

$\frac{d}{dx} ( \sec(x) ) = \tan(x) \sec(x)$

$\frac{d}{dx} ( log(x) ) = \frac{1}{x}$