What is the tangent of f(x)=x^2-sqrt(x),\at x=4 ?

The tangent of f(x)=x^2-sqrt(x),\at x=4 is y= 31/4 x-17

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tangent of f(x)=x^2-sqrt(x),\at x=4

tangent of

f (x) = x^{2} - x, at x = 4

y = \frac{31}{4} x - 17

Solution steps

Find the tangent point:

(4, 14)

Find the slope of

f (x) = x^{2} - x : \frac{df ( x )}{d x} = 2 x - \frac{1}{2 x}

Compute the slope of

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

x = 4 : m = \frac{31}{4}

Find the line with slope m=

\frac{31}{4}

and passing through

(4, 14) : y = \frac{31}{4} x - 17

y = \frac{31}{4} x - 17

y^{^{'}} = y ln (\frac{450}{y})

f (x) = 4 cos (4 x)

\int \frac{1}{e ^{- y} + 1} d y

\frac{d}{d x} (\frac{x}{1 - ln ( x - 6 )})

\frac{\partial}{\partial y} (\frac{x}{y + z})

What is the tangent of f(x)=x^2-sqrt(x),\at x=4 ?
The tangent of f(x)=x^2-sqrt(x),\at x=4 is y= 31/4 x-17