What is the tangent of y=x^{sqrt(x)},(4,16) ?

The tangent of y=x^{sqrt(x)},(4,16) is y=8(ln(2)+1)x-32ln(2)-16

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tangent of y=x^{sqrt(x)},(4,16)

tangent of

y = x^{x}, (4, 16)

y = 8 (ln (2) + 1) x - 32 ln (2) - 16

Solution steps

Find the slope of

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

Compute the slope of

y = x^{x}

(4, 16) : m = 8 (ln (2) + 1)

Find the line with slope m=

8 (ln (2) + 1)

and passing through

(4, 16) : y = 8 (ln (2) + 1) x - 32 ln (2) - 16

y = 8 (ln (2) + 1) x - 32 ln (2) - 16

\frac{d}{d x} (e^{l n (x^{2})})

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

4^{x} - 2^{x}

\frac{\partial}{\partial x} (2 x^{2} + 9 y^{2} + 3)

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

What is the tangent of y=x^{sqrt(x)},(4,16) ?
The tangent of y=x^{sqrt(x)},(4,16) is y=8(ln(2)+1)x-32ln(2)-16