# How do you implicitly differentiate #7xy- 3 lny= 4/x#?

Find the derivative of each part.

Use product rule:

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To implicitly differentiate (7xy - 3\ln(y) = \frac{4}{x}), follow these steps:

- Differentiate each term with respect to (x).
- Apply the chain rule when differentiating (\ln(y)).

Differentiating each term:

[\frac{d}{dx}(7xy) - \frac{d}{dx}(3\ln(y)) = \frac{d}{dx}\left(\frac{4}{x}\right)]

Apply product rule to (7xy):

[7\frac{d}{dx}(xy) + x\frac{d}{dx}(7y) - 3\frac{d}{dx}(\ln(y)) = \frac{d}{dx}\left(\frac{4}{x}\right)]

Apply chain rule to (\ln(y)):

[7\left(\frac{dy}{dx}y + x\frac{dy}{dx}\right) - 3\left(\frac{1}{y}\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{4}{x}\right)]

Differentiate (\frac{4}{x}):

[-\frac{4}{x^2} = -\frac{4}{x^2}]

Simplify and solve for (\frac{dy}{dx}):

[7xy' + 7y + xy' - \frac{3}{y}y' = -\frac{4}{x^2}]

[7xy' + xy' - \frac{3}{y}y' = -\frac{4}{x^2} - 7y]

[xy'(7 + 1) - \frac{3}{y}y' = -\frac{4}{x^2} - 7y]

[8xy' - \frac{3}{y}y' = -\frac{4}{x^2} - 7y]

[y'(8x - \frac{3}{y}) = -\frac{4}{x^2} - 7y]

[y' = \frac{-\frac{4}{x^2} - 7y}{8x - \frac{3}{y}}]

This is the implicit derivative of the given function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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