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}

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{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} f (x) = \frac{12}{( 3 x - 4 ) ^{5}}

\int \frac{- 2}{z ^{3}} d z

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + e^{- x}

\frac{d}{d x} (\frac{sin ( x )}{1 + sin ( x )})

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