What is the tangent of y=1+sqrt(x),(4,3) ?

The tangent of y=1+sqrt(x),(4,3) is y= 1/4 x+2

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tangent of y=1+sqrt(x),(4,3)

tangent of

y = 1 + x, (4, 3)

y = \frac{1}{4} x + 2

Solution steps

Find the slope of

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

Compute the slope of

y = 1 + x

(4, 3) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(4, 3) : y = \frac{1}{4} x + 2

y = \frac{1}{4} x + 2

\frac{d}{d x} (- \frac{1}{( 3 + x ) ^{2}})

x \to - 3 lim (\frac{x ^{3}}{x ^{2} - 9})

\frac{d}{d x} (sin (h) (x))

f (x) = e^{x}

\frac{d}{d x} (sin^{1} (x))

What is the tangent of y=1+sqrt(x),(4,3) ?
The tangent of y=1+sqrt(x),(4,3) is y= 1/4 x+2