What is the tangent of g(x)=sqrt(x),\at x=121 ?

The tangent of g(x)=sqrt(x),\at x=121 is y= 1/22 x+11/2

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tangent of g(x)=sqrt(x),\at x=121

tangent of

g (x) = x, at x = 121

y = \frac{1}{22} x + \frac{11}{2}

Solution steps

Find the tangent point:

(121, 11)

Find the slope of

g (x) = x : \frac{d g ( x )}{d x} = \frac{1}{2 x}

Compute the slope of

g (x) = x

x = 121 : m = \frac{1}{22}

Find the line with slope m=

\frac{1}{22}

and passing through

(121, 11) : y = \frac{1}{22} x + \frac{11}{2}

y = \frac{1}{22} x + \frac{11}{2}

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

\int_{0}^{1} 5 (1 + x)^{5} d x

\int x ln (x^{2}) d x

x \to 2 lim (k x^{2})

\frac{d}{d x} (\frac{d y}{d x}) + 5 y = 7

What is the tangent of g(x)=sqrt(x),\at x=121 ?
The tangent of g(x)=sqrt(x),\at x=121 is y= 1/22 x+11/2