What is the tangent of f(x)=sqrt(x^2+13),\at x=6 ?

The tangent of f(x)=sqrt(x^2+13),\at x=6 is y= 6/7 x+13/7

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tangent of f(x)=sqrt(x^2+13),\at x=6

tangent of

f (x) = x^{2} + 13, at x = 6

y = \frac{6}{7} x + \frac{13}{7}

Solution steps

Find the tangent point:

(6, 7)

Find the slope of

f (x) = x^{2} + 13 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 13}

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{6}{7}

Find the line with slope m=

\frac{6}{7}

and passing through

(6, 7) : y = \frac{6}{7} x + \frac{13}{7}

y = \frac{6}{7} x + \frac{13}{7}

\int 5^{1 - x} d x

\frac{d y}{d x} = 2 y + 34 x + 52

x \to 0 + lim (lo g_{\frac{1}{e}} (x))

(2 y^{2} + 3 x y) + (2 x y + x^{2}) \frac{d y}{d x} = 0

\int_{1}^{3} x^{2} e^{x^{3} - 1} d x

What is the tangent of f(x)=sqrt(x^2+13),\at x=6 ?
The tangent of f(x)=sqrt(x^2+13),\at x=6 is y= 6/7 x+13/7