Question

x= e^cos3t y=e^sin3t then dy/dx​

Answer 1

Given :

$x= {e}^{cos3t}$

$y=e^{sin3t}$

To find :

$\dfrac{dy}{dx}$

Concept :

$\dfrac{dy}{dx} = \dfrac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\dfrac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \dfrac{dy \times dt }{dx \times dt}$

dt will get cancelled from numerator and denominator :-

$\dfrac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \dfrac{dy }{dx}$

Solution :

Now first we will find dy/dt :-

$\dfrac{d(y)}{dt} = \dfrac{d(e^{sin3t})}{dt}$

we know :-

• d (e^x)/dx = e^x dx/dx.
• d (sin x)/dx = cos x dx/dx.

By applying Chain rule :-

$\dfrac{d(y)}{dt} =e^{sin3t} \times cos \: 3t \: \times 3$

Now , we will find dx/dt :-

$\dfrac{d(x)}{dt} = \dfrac{d({e}^{cos3t})}{dt}$

we know :-

• d (e^x)/dx = e^x dx/dx.
• d (cos x)/dx = - sin x dx/dx.

By applying Chain rule :-

$\dfrac{d(x)}{dt} = {e}^{cos3t} \times sin \: 3t \times 3$

Now we will find dy/dx :-

$\dfrac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \dfrac{e^{sin3t} \times cos \: 3t \: \times 3}{ {e}^{cos3t} \times sin \: 3t \times 3}$

( cos x/ sinx = cotx)

$\dfrac{dy}{dx} = e^{sin3t - cos3t} cot \:3t$

ANSWER :

$\dfrac{dy}{dx} = e^{sin3t - cos3t} cot \: 3t$

____________________

Learn more :-

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

d(log x )/dx= 1/x

d (e^x)/dx = e^x

________________________