What is the tangent of y=(5x)/(2^x),\at x=2 ?

The tangent of y=(5x)/(2^x),\at x=2 is y=(5(1-2ln(2)))/4 x+5ln(2)

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tangent of y=(5x)/(2^x),\at x=2

tangent of

y = \frac{5 x}{2 ^{x}}, at x = 2

y = \frac{5 ( 1 - 2 ln ( 2 ) )}{4} x + 5 ln (2)

Solution steps

Find the tangent point:

(2, \frac{5}{2})

Find the slope of

y = \frac{5 x}{2 ^{x}} : \frac{d y}{d x} = \frac{5 ( 1 - ln ( 2 ) x )}{2 ^{x}}

Compute the slope of

y = \frac{5 x}{2 ^{x}}

x = 2 : m = \frac{5 ( 1 - 2 ln ( 2 ) )}{4}

Find the line with slope m=

\frac{5 ( 1 - 2 ln ( 2 ) )}{4}

and passing through

(2, \frac{5}{2}) : y = \frac{5 ( 1 - 2 ln ( 2 ) )}{4} x + 5 ln (2)

y = \frac{5 ( 1 - 2 ln ( 2 ) )}{4} x + 5 ln (2)

\frac{\partial}{\partial y} (ln (x^{2} + 3 y))

\frac{d}{d x} (\frac{6 x ^{5} - 7 x}{x ^{4} - 8})

\int (6 e^{- 0.2 x}) d x

5 - x, 25 - x^{2}

\int \frac{1}{x ^{2}} arctan (x) d x

What is the tangent of y=(5x)/(2^x),\at x=2 ?
The tangent of y=(5x)/(2^x),\at x=2 is y=(5(1-2ln(2)))/4 x+5ln(2)