# 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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