Parametric Equations. Find $\frac{dy}{dx}$ in terms of $x$

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Find $\frac{dy}{dx}$ in terms of $x$ if the parametric equations of a curve are given by $x=e^{\sqrt{4t}}$ and $y=\sqrt{e^{6t}}$.

My attempt,

I found $\frac{dx}{dt}=\frac{e^{2\sqrt{t}}}{\sqrt{t}}$ and $\frac{dy}{dt}=3e^{3t}$. How to find $\frac{dy}{dx}$ in terms of $x$?

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2 Answers

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HINT:

$$\ln x=\sqrt{4t}\implies4t=(\ln x)^2$$

and $$\dfrac12\ln y=6t\iff \ln y=12t=3(\ln x)^2$$

Now differentiate wrt $x$

and use $y=e^{3(\ln x)^2}=(e^{\ln x})^{3\ln x}=x^{3\ln x}$

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Formula....$dy/dx=(dy/dt)/(dx/dt) $

To find the answer in terms of x.. find the inverse function of x(t) and then substitute t as a function of x with the help of inverse function.....

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