Question

Q. 11 Find
if x = cos (logt), y = log (cost)​

Answer 1

Answer:-

$\bf{ \frac{dy}{dx} = \frac{t .\: tan \: t}{sin \: (log \: t)} } \:$

Step - by - step explanation:-

Given :-

→ x = cos ( log t), y = log (cos t)

To find :-

Find dy/dx

Solution:-

Firstly ,find derivative of x with respect to "t" .

• x = cos ( log t)

differentiate to X with respect to "t "

$\bf{ \frac{dx}{dt} = - sin \: (log \:t) \frac{d}{dt} log \: t} \\ \\ \bf{ \frac{dx}{dt} = - sin \: (log \: t) \times \frac{1}{t} } \\ \\ \bf{ \frac{dx}{dt} = \frac{ - sin \: (log \: t)}{t} } \: ........(1)$

Now ,

Find derivative of y with respect to "t "

• y = log ( cos t)

differentiate y with respect to "t"

$\bf{\frac{dy}{dt} = \frac{1}{cos \: t} \frac{d}{dt} (cos \: t)} \\ \\ \bf{\frac{dy}{dt} = - \frac{sin \: t}{cos \: t} } \\ \\ \bf{ \frac{dy}{dt} = - tan \: t} \: ...........(2)$

Now ,divide equation (2) by (1)

Then we get ,

$\bf{ \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{ - tan \: t}{ \frac{ - sin \: (log \: t)}{t} } } \\ \\ \bf{ \frac{dy}{dx} = \frac{t .\: tan \: t}{sin \: (log \: t)} }$

Hope it helps you.

Answer 2

Answer:

dy/dx=t.tant/sin(log)

Step-by-step explanation:

Consider,

x=cos(logt)

Diff.w.r.t.t

dx/dt =-sin(logt) d/dt log

dx/dr=-sin(logt)×1/t

dx/dt=-sin(logt)/t.......1

Also,

y=log(cost)

Diff.w.r.t.t

dy/dt=1/cost d/dt(cost)

dy/dt=-sint/cost ....(sin/cos=tan)

dy/dt=-tant..........2

Now,

dy/dx= dy/dy /dx/dt= -tant/-sin(logt)/t Therefore,

dy/dx=t.tant/sin(logt)

Thank you...