Question

Find the derivatives of the following functions w.r.to x ( i ) y = sec (log x )​

Answer 1

Answer:

y=sec(logx)

dy/dx=d/dx[sec(logx)]

dy/dx=sec(logx)tan(logx).d/dx(logx)

dy/dx=[sec(logx).tan(logx)]/x

Note that derivative of secx is secxtanx and that of logx is 1/x.

Answer 2

Answer:

The derivative of the given function is

$\displaystyle{\boxed{\red{\sf\:\dfrac{\sec\:(\:\log\:x\:)\:\tan\:(\:\log\:x\:)}{x}}}}$

Step-by-step-explanation:

We have given a function.

We have to find its derivative w. r. t. x.

The given function is

$\displaystyle{\sf\:y\:=\:\sec\:(\:\log\:x\:)}$

Let $\displaystyle{\sf\:\log\:x\:=\:u}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{dy}{dx}\:[\:f\:(\:g\:(\:x\:))\:]\:=\:f'\:(\:g\:(\:x\:))\:g'\:(\:x\:)}}}$

Now,

$\displaystyle{\sf\:y\:=\:\sec\:u}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{d\:(\:\sec\:u\:)}{dx}\:.\:\dfrac{dy}{dx}\:u}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\sec\:u\:\tan\:u\:.\:\dfrac{d\:(\:\log\:x\:)}{dx}\:\quad\:\dots\:[\:u\:=\:\log\:x\:]}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\sec\:(\:\log\:x\:)\:\tan\:(\:\log\:x\:)\:\dfrac{1}{x}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{\sec\:(\:\log\:x\:)\:\tan\:(\:\log\:x\:)}{x}}}}}$