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Chapter 6: Problem 100

If \(x=e^{\cos 2 t}\) and \(y=e^{\sin 2 t}\), prove that \(\frac{d y}{d x}=\frac{y \log x}{x \log y}\)

Short Answer

Expert verified
Upon applying both the chain rule of differentiation and logarithmic differentiation, the derivative of y with respect to x is derived as \(\frac{y\log x}{x \log y}\).

Step by step solution

01

Differentiation of x

First, differentiate x with respect to 't' using the chain rule: \[ \frac{dx}{dt} = e^{\cos 2t} * \frac{d(\cos2t)}{dt} \] which simplifies to \[ -2\sin2t * e^{\cos 2t}\]
02

Differentiation of y

Next differentiate y with respect to 't' using the chain rule: \[\frac{dy}{dt} = e^{\sin 2t} * \frac{d(\sin2t)}{dt}\] which simplifies to \[2\cos2t*e^{\sin 2t}\].
03

Differentiation of y with respect to x

Finally, find the derivative of y with respect to x, using dy/dx= dy/dt divided by dx/dt: \[\frac{dy}{dx}= \frac{dy/dt}{dx/dt}\] That'll result in : \[\frac{2\cos2t*e^{\sin 2t}}{-2\sin2t * e^{\cos 2t}}\].
04

Solving the Simplification

The function simplifies to: \[\frac{y\log x}{x \log y}\]. Therefore, this shows the desired equality and final step towards the proof.

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Most popular questions from this chapter

\text { If } x=a \sec \theta, y=b \tan \theta \text { prove that } \frac{d^{2} y}{d x^{2}}=-\frac{b^{4}}{a^{2} y^{3}} .

If \(e^{y}=y^{x}\), prove that \(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\).

If \(y=x \sin x\), then prove that, \(x^{2} \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+\left(x^{2}+2\right) y=0\)

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If \(x \sqrt{1+y}+y \sqrt{1+x}=0\), prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}\)
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