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

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

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

tangent of

f (x) = 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

f (x) = x^{2} + 21 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 21}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

\frac{\partial}{\partial z} (x^{2} + y^{2})

\frac{d y}{d x} = \frac{- 2 x}{y}

y = x + x

(x + y e^{(\frac{y}{x})}) d x - (x e^{\frac{y}{x}}) d y = 0

\frac{d y}{e ^{y}} = x d x

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