What is the tangent of sqrt(x^2+16),\at x=3 ?

The tangent of sqrt(x^2+16),\at x=3 is y= 3/5 x+16/5

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tangent of sqrt(x^2+16),\at x=3

tangent of

x^{2} + 16, at x = 3

y = \frac{3}{5} x + \frac{16}{5}

Solution steps

Find the tangent point:

(3, 5)

Find the slope of

y = x^{2} + 16 : \frac{d y}{d x} = \frac{x}{x ^{2} + 16}

Compute the slope of

y = x^{2} + 16

x = 3 : m = \frac{3}{5}

Find the line with slope m=

\frac{3}{5}

and passing through

(3, 5) : y = \frac{3}{5} x + \frac{16}{5}

y = \frac{3}{5} x + \frac{16}{5}

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{d}{d x} (2 x^{2} + 5)

y = cot^{2} (cos (x))

\frac{d}{d t} (\frac{t}{t - 2})

\int \frac{- 1 - u}{u ^{3} + u ^{2} - 8 u - 12} d u

What is the tangent of sqrt(x^2+16),\at x=3 ?
The tangent of sqrt(x^2+16),\at x=3 is y= 3/5 x+16/5