What is the tangent of y=sqrt(x),(49,7) ?

The tangent of y=sqrt(x),(49,7) is y= 1/14 x+7/2

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tangent of y=sqrt(x),(49,7)

tangent of

y = x, (49, 7)

y = \frac{1}{14} x + \frac{7}{2}

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

(49, 7) : m = \frac{1}{14}

Find the line with slope m=

\frac{1}{14}

and passing through

(49, 7) : y = \frac{1}{14} x + \frac{7}{2}

y = \frac{1}{14} x + \frac{7}{2}

\int \frac{25}{x ^{\frac{3}{2}}} d x

\int x \cdot 2^{x^{2}} d x

f^{'} (x)

\int x sec^{2} (4 x) d x

f (x) = (x^{2} - 15)^{7}, at x = 4

What is the tangent of y=sqrt(x),(49,7) ?
The tangent of y=sqrt(x),(49,7) is y= 1/14 x+7/2