How do you use implicit differentiation to find the slope of the line tangent to $y + ln x y = 4$ $y+ lnxy=4$ at (.25, 4)?

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How do you use implicit differentiation to find the slope of the line tangent to $y + ln x y = 4$ $y+ lnxy=4$ at (.25, 4)?

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Answer 1 · 2018-03-21T23:02:18

$\frac{d y}{d x} = - \frac{16}{5}$ $dy/dx=-16/5$ = $[3.2$ $[3.2$ ] [The gradient]

Explanation:

$y + ln x y = 4$ $y+lnxy=4$ , so by the theory of logs, $y + ln x + ln y = 4$ $y+ lnx+lny=4$

Differentiating both sides of the equation implicitly with respect to x,....... $\frac{d y}{d x} + \frac{1}{x} + \frac{1}{y} \frac{d y}{d x} = 0$ $dy/dx+1/x+1/ydy/dx=0$ , Therefore,

$\frac{d y}{d x} [1 + \frac{1}{y}] = - \frac{1}{x}$ $dy/dx[1+1/y]=-1/x$ ...... so $\frac{d y}{d x} = - \frac{y}{x [y + 1]}$ $dy/dx=-y/[x[y+1]]$ and substituting for $x = 0.25 and y = 4$ $x=0.25 and y=4$ will give the above result for the gradient.

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How do you use implicit differentiation to find the slope of the line tangent to $y + ln x y = 4$ $y+ lnxy=4$ at (.25, 4)?

1 Answer

Explanation:

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How do you use implicit differentiation to find the slope of the line tangent to y+ lnxy=4y+lnxy=4y+ lnxy=4 at (.25, 4)?

1 Answer

Explanation:

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How do you use implicit differentiation to find the slope of the line tangent to $y + ln x y = 4$ $y+ lnxy=4$ at (.25, 4)?