Question #829b0

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Question #829b0

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Answer 1 · 2018-02-01T23:43:44

Given $ln (5 x - 3 y^{2})$ $ln(5x-3y^2)$ , $\frac{d}{d x} [ln (5 x - 3 y^{2})] = \frac{5 - 6 y}{5 x - 3 y^{2}}$ $d/dx[ln(5x-3y^2)]=(5-6y)/(5x-3y^2)$

Explanation:

$ln (5 x - 3 y^{2})$ $ln(5x-3y^2)$
First, you derive the inside of the function ( $5 x - 3 y^{2}$ $5x-3y^2$ )
1. $5 - 6 y$ $5-6y$
Second, you put this answer over the inside of the function.
2. $\frac{d}{d x} [ln (5 x - 3 y^{2})] = \frac{5 - 6 y}{5 x - 3 y^{2}}$ $d/dx[ln(5x-3y^2)]=(5-6y)/(5x-3y^2)$ .

Answer 2 · 2018-02-01T23:48:14

Please see below.

Explanation:

.

Let $u = 5 x - 3 y^{2}$ $u=5x-3y^2$

$d u = 5 d x - 6 y d y$ $du=5dx-6ydy$

$d (ln (5 x - 3 y^{2})) = d (ln u) = \frac{1}{u} d u = \frac{5 d x - 6 y d y}{5 x - 3 y^{2}}$ $d(ln(5x-3y^2))=d(lnu)=1/udu=(5dx-6ydy)/(5x-3y^2)$

If we only differentiate with respect to $x$ $x$ we will have:

$\frac{d u}{d x} = 5$ $(du)/dx=5$

$\frac{d}{d x} (ln (5 x - 3 y^{2})) = \frac{d}{d x} (ln u) = \frac{1}{u} d u = \frac{5}{5 x - 3 y^{2}}$ $d/dx(ln(5x-3y^2))=d/dx(lnu)= 1/udu=5/(5x-3y^2)$

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Question #829b0

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