What is the tangent of f(x)=sqrt(x),\at x=16.3 ?

The tangent of f(x)=sqrt(x),\at x=16.3 is y=0.12384…x+2.01866…

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tangent of f(x)=sqrt(x),\at x=16.3

tangent of

f (x) = x, at x = 16.3

y = 0.12384 \dots x + 2.01866 \dots

Solution steps

Find the tangent point:

(16.3, 16.3)

Find the slope of

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

Compute the slope of

f (x) = x

x = 16.3 : m = 0.12384 \dots

Find the line with slope m=

0.12384 \dots

and passing through

(16.3, 16.3) : y = 0.12384 \dots x + 2.01866 \dots

y = 0.12384 \dots x + 2.01866 \dots

arcsin (x)

\frac{d ^{2}}{d x ^{2}} (2 x cos (x^{2}))

\int_{8}^{14} \frac{y}{y ^{2} - 4 y - 5} d y

y^{^{'}} = e^{- 2 x}

26 e^{x^{2}} x

What is the tangent of f(x)=sqrt(x),\at x=16.3 ?
The tangent of f(x)=sqrt(x),\at x=16.3 is y=0.12384…x+2.01866…