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I have this old examination assignment, where I have a function for some curve and the coordinates to a point. The subject is to determine the degree of the curve in the given point.

The expression of the curve is: $$ x^2 \sin{\sqrt{y}}+ye^{-2x}=1 $$ and the point is (0,1). The first thing I should do is to implicit differentiate the function. I am not really sure if I have done that right, here is my answer:

$$ 2x \cos{\sqrt{y}}\times\frac{1}{2}y^{-\frac{1}{2}}\times y'+y\times y'(-2)e^{-2x}=0$$

Is this right?

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2 Answers 2

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You need to use the product rule to differentiate, implicitly, $$x^2 \sin{\sqrt{y}}$$ and $$ye^{-2x}$$

recall: when you have the product of two functions $f(x)g(x)$, $$(f(x)g(x))' = f'(x)g(x) + f(x) g'(x)$$

So, for example, to differentiate $x^2 \sin{\sqrt y}$, we obtain:

$$2x\sin{\sqrt{y}}+x^2\cos({\sqrt y})\frac{y'}{2\sqrt{y}}\tag{1}$$

And to differentiate $ye^{-2x}$, we obtain: $$y'e^{-2x} + y(-2)e^{-2x} = y'e^{-2x}- 2ye^{-2x}\tag{2}$$

Now sum $(1), (2)$ and differentiate the the right-hand side of the original equation $(1)' = 0$ to obtain your derivative.

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  • $\begingroup$ See here $\endgroup$
    – Mikasa
    Jan 25, 2014 at 14:51
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$$x^2 \sin{\sqrt{y}}+ye^{-2x}=1\to\color{red}{(x^2 \sin{\sqrt{y}})_x}+\color{blue}{(ye^{-2x})_x}=1_x=0$$ in which $~~\cdot_x~~$ means the drivation with respect to $x$. We have then: $$\color{red}{\left(2x\sin{\sqrt{y}}+x^2\cos{\sqrt{y}}\frac{y'}{2\sqrt{y}}\right)}+\color{blue}{(y'e^{-2x}-2ye^{-2x})}=0$$

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