How do you find the equation of the tangent line to the graph $y = 2^{- x}$ $y=2^-x$ through point (-1,2)?

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How do you find the equation of the tangent line to the graph $y = 2^{- x}$ $y=2^-x$ through point (-1,2)?

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Answer 1 · 2017-04-20T18:14:57

Use logarithmic differentiation.
Evaluate the derivative at $x = - 1$ $x = -1$ , to find the slope.
Use the point-slope form of the equation of the line.

Explanation:

Use the natural logarithm on both sides of the equation:

$ln (y) = ln (2^{- x})$ $ln(y)=ln(2^-x)$

Use the property $ln (a^{c}) = (c) ln (a)$ $ln(a^c) = (c)ln(a)$

$ln (y) = - x ln (2) = ln (\frac{1}{2}) x$ $ln(y)=-xln(2) = ln(1/2)x$

Differentiate both sides:

$\frac{1}{y} \frac{d y}{d x} = ln (\frac{1}{2})$ $1/ydy/dx=ln(1/2)$

Multiply both sides by y:

$\frac{d y}{d x} = ln (\frac{1}{2}) y$ $dy/dx=ln(1/2)y$

Substitute for y:

$\frac{d y}{d x} = ln (\frac{1}{2}) 2^{- x}$ $dy/dx=ln(1/2)2^-x$

The slope, m, of the tangent line is the derivative evaluated at $x = - 1$ $x=-1$ :

$m = ln (\frac{1}{2}) 2^{- (- 1)}$ $m=ln(1/2)2^-(-1)$

$m = 2 ln (\frac{1}{2})$ $m=2ln(1/2)$

$m = ln (\frac{1}{4})$ $m=ln(1/4)$

Using the point-slope form of the equation of the line:

$y = m (x - x_{1}) + y_{1}$ $y = m(x-x_1)+y_1$

$y = ln (\frac{1}{4}) (x - - 1) + 2$ $y = ln(1/4)(x--1)+2$

$y = ln (\frac{1}{4}) (x + 1) + 2$ $y = ln(1/4)(x+1)+2$

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How do you find the equation of the tangent line to the graph $y = 2^{- x}$ $y=2^-x$ through point (-1,2)?

1 Answer

Explanation:

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How do you find the equation of the tangent line to the graph $y = 2^{- x}$ $y=2^-x$ through point (-1,2)?