Find [tex]\frac{d y}{d x}[/tex] by implicit differentiation.

[tex]\ \textless \ br/\ \textgreater \ \sqrt{xy} = x^8 y + 60\ \textless \ br/\ \textgreater \ [/tex]

[tex]\ \textless \ br/\ \textgreater \ \frac{d y}{d x} = \square\ \textless \ br/\ \textgreater \ [/tex]



Answer :

To find [tex]\(\frac{dy}{dx}\)[/tex] by implicit differentiation, we will differentiate both sides of the equation with respect to [tex]\(x\)[/tex]. Let's go through the steps together:

First, consider the given equation:
[tex]\[ \sqrt{xy} = x^8 y + 60 \][/tex]

### Step 1: Differentiate both sides with respect to [tex]\(x\)[/tex]
Differentiate the left-hand side:
[tex]\[ \frac{d}{dx} \left( \sqrt{xy} \right) \][/tex]

Using the chain rule on [tex]\(\sqrt{xy}\)[/tex]:
[tex]\[ \frac{d}{dx} \left( xy \right)^{1/2} = \frac{1}{2} (xy)^{-1/2} \cdot \frac{d}{dx} (xy) \][/tex]

Now, apply the product rule to [tex]\(\frac{d}{dx} (xy)\)[/tex]:
[tex]\[ \frac{d}{dx} (xy) = x \frac{d y}{dx} + y \][/tex]

Putting it together:
[tex]\[ \frac{d}{dx} (xy)^{1/2} = \frac{1}{2} (xy)^{-1/2} \left( x \frac{d y}{dx} + y \right) \][/tex]

So, the derivative of the left-hand side is:
[tex]\[ \frac{1}{2} \frac{x y^{1/2}}{(xy)^{1/2}} \left( x \frac{d y}{dx} + y \right) \][/tex]

Since [tex]\(\sqrt{xy} = (xy)^{1/2}\)[/tex], this simplifies to:
[tex]\[ \frac{1}{2 \sqrt{xy}} (x \frac{d y}{dx} + y) \][/tex]

Now, differentiate the right-hand side:
[tex]\[ \frac{d}{dx} (x^8 y + 60) \][/tex]

Using the product rule on [tex]\(x^8 y\)[/tex]:
[tex]\[ = 8x^7 y + x^8 \frac{d y}{dx} \][/tex]

And the derivative of the constant 60 is 0.

So, the derivative of the right-hand side is:
[tex]\[ 8 x^7 y + x^8 \frac{d y}{dx} \][/tex]

### Step 2: Equate the derivatives and solve for [tex]\(\frac{dy}{dx}\)[/tex]
Equate the derivatives from both sides:
[tex]\[ \frac{1}{2 \sqrt{xy}} (x \frac{d y}{dx} + y) = 8x^7 y + x^8 \frac{d y}{dx} \][/tex]

Multiply through by [tex]\(2 \sqrt{xy}\)[/tex] to clear the denominator:
[tex]\[ x \frac{d y}{dx} + y = 16 x^7 y \sqrt{xy} + 2 x^8 \frac{d y}{dx} \sqrt{xy} \][/tex]

Collect all terms involving [tex]\(\frac{d y}{dx}\)[/tex] on one side:
[tex]\[ x \frac{d y}{dx} - 2 x^8 \frac{d y}{dx} \sqrt{xy} = 16 x^7 y \sqrt{xy} - y \][/tex]

Factor out [tex]\(\frac{d y}{dx}\)[/tex]:
[tex]\[ \frac{d y}{dx} (x - 2 x^8 \sqrt{xy}) = 16 x^7 y \sqrt{xy} - y \][/tex]

Finally, solve for [tex]\(\frac{d y}{dx}\)[/tex]:
[tex]\[ \frac{d y}{dx} = \frac{16 x^7 y \sqrt{xy} - y}{x - 2 x^8 \sqrt{xy}} \][/tex]

This simplifies to:
[tex]\[ \frac{d y}{dx} = -8 x^7 y + \frac{\sqrt{xy}}{2 x} \][/tex]

Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \frac{dy}{dx} = -8 x^7 y + \frac{\sqrt{xy}}{2 x} \][/tex]

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