What is the tangent of y=9x-8sqrt(x),(1,1) ?

The tangent of y=9x-8sqrt(x),(1,1) is y=5x-4

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tangent of y=9x-8sqrt(x),(1,1)

tangent of

y = 9 x - 8 x, (1, 1)

y = 5 x - 4

Solution steps

Find the slope of

y = 9 x - 8 x : \frac{d y}{d x} = 9 - \frac{4}{x}

Compute the slope of

y = 9 x - 8 x

(1, 1) : m = 5

Find the line with slope m=

5

and passing through

(1, 1) : y = 5 x - 4

y = 5 x - 4

\frac{( 2 - z ) ^{\frac{2}{3}}}{4 z + 5}

\frac{d}{d x} ((3 x + 2)^{4})

f (x) = 28 x^{2} - 10 x - 2

\frac{d y}{d x} = sin (y) ln (x)

\int \frac{2 x}{x ^{2} + 4 x + 4} d x

What is the tangent of y=9x-8sqrt(x),(1,1) ?
The tangent of y=9x-8sqrt(x),(1,1) is y=5x-4