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

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

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

tangent of

f (x) = 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

f (x) = x^{2} + 16 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 16}

Compute the slope of

f (x) = 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 sin^{3} (5 x) cot (5 x) d x

6 y^{^{''}} + 9 y^{^{'}} = 0

\int \frac{1}{( x ^{2} - 2 x + 1 ) x ^{2} - 2 x} d x

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

\frac{d}{d x} (2 x^{5} + 6 x^{8})

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