What is the tangent of f(x)=xsqrt(x^2+60),\at x=2 ?

The tangent of f(x)=xsqrt(x^2+60),\at x=2 is y= 17/2 x-1

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tangent of f(x)=xsqrt(x^2+60),\at x=2

tangent of

f (x) = x x^{2} + 60, at x = 2

y = \frac{17}{2} x - 1

Solution steps

Find the tangent point:

(2, 16)

Find the slope of

f (x) = x x^{2} + 60 : \frac{df ( x )}{d x} = \frac{2 x ^{2} + 60}{x ^{2} + 60}

Compute the slope of

f (x) = x x^{2} + 60

x = 2 : m = \frac{17}{2}

Find the line with slope m=

\frac{17}{2}

and passing through

(2, 16) : y = \frac{17}{2} x - 1

y = \frac{17}{2} x - 1

\frac{\partial}{\partial x} ((3 - x) (3 - y) (x + y - 3))

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 3 x})

x (- 3 x^{- 4})

\int - 5 sin^{3} (x) cos^{2} (x) d x

\frac{d y}{d t} = 1 0^{- 21} \cdot y^{3}

What is the tangent of f(x)=xsqrt(x^2+60),\at x=2 ?
The tangent of f(x)=xsqrt(x^2+60),\at x=2 is y= 17/2 x-1