What is the tangent of f(x)=xsqrt(x),\at x=1 ?

The tangent of f(x)=xsqrt(x),\at x=1 is y= 3/2 x-1/2

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tangent of f(x)=xsqrt(x),\at x=1

tangent of

f (x) = x x, at x = 1

y = \frac{3}{2} x - \frac{1}{2}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = x x : \frac{df ( x )}{d x} = \frac{3 x}{2}

Compute the slope of

f (x) = x x

x = 1 : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 1) : y = \frac{3}{2} x - \frac{1}{2}

y = \frac{3}{2} x - \frac{1}{2}

y^{^{''}} + \frac{1}{x} y^{^{'}} - \frac{1}{x ^{2}} y = \frac{2}{x}

y = \frac{e ^{x} + e ^{- x}}{2}

f (x) = e^{1 - x}

\int x^{2} arctan (7 x) d x

\int \frac{1}{1 - 4 x ^{2}} d x

What is the tangent of f(x)=xsqrt(x),\at x=1 ?
The tangent of f(x)=xsqrt(x),\at x=1 is y= 3/2 x-1/2