What is the tangent of f(x)=sqrt(x^2+33),\at x=4 ?

The tangent of f(x)=sqrt(x^2+33),\at x=4 is y= 4/7 x+33/7

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tangent of f(x)=sqrt(x^2+33),\at x=4

tangent of

f (x) = x^{2} + 33, at x = 4

y = \frac{4}{7} x + \frac{33}{7}

Solution steps

Find the tangent point:

(4, 7)

Find the slope of

f (x) = x^{2} + 33 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 33}

Compute the slope of

f (x) = x^{2} + 33

x = 4 : m = \frac{4}{7}

Find the line with slope m=

\frac{4}{7}

and passing through

(4, 7) : y = \frac{4}{7} x + \frac{33}{7}

y = \frac{4}{7} x + \frac{33}{7}

\frac{\partial}{\partial x} (2 x + 4 y^{2})

\int (x + 1) (x - 1) d x

\int \frac{x ^{2} - 5 x + 1}{3 x ^{2}} d x

\frac{\partial}{\partial x} (2 sin (5 x) cos (4 y))

\int sin^{2} (t) cos^{2} (t) d t

What is the tangent of f(x)=sqrt(x^2+33),\at x=4 ?
The tangent of f(x)=sqrt(x^2+33),\at x=4 is y= 4/7 x+33/7