What is the tangent of sqrt(x^2+21),\at x=2 ?

The tangent of sqrt(x^2+21),\at x=2 is y= 2/5 x+21/5

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tangent of sqrt(x^2+21),\at x=2

tangent of

x^{2} + 21, at x = 2

y = \frac{2}{5} x + \frac{21}{5}

Solution steps

Find the tangent point:

(2, 5)

Find the slope of

y = x^{2} + 21 : \frac{d y}{d x} = \frac{x}{x ^{2} + 21}

Compute the slope of

y = x^{2} + 21

x = 2 : m = \frac{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(2, 5) : y = \frac{2}{5} x + \frac{21}{5}

y = \frac{2}{5} x + \frac{21}{5}

x \to 1 lim (\frac{4 x - 4}{x - 1})

\int 16 x sin (x) d x

\frac{1 + x}{1 - x}

\int_{0}^{e} ln (x) d x

\int (1 + e)^{x} d x

What is the tangent of sqrt(x^2+21),\at x=2 ?
The tangent of sqrt(x^2+21),\at x=2 is y= 2/5 x+21/5