What is the tangent of f(x)=sqrt(x^2+24),\at x=5 ?

The tangent of f(x)=sqrt(x^2+24),\at x=5 is y= 5/7 x+24/7

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tangent of f(x)=sqrt(x^2+24),\at x=5

tangent of

f (x) = x^{2} + 24, at x = 5

y = \frac{5}{7} x + \frac{24}{7}

Solution steps

Find the tangent point:

(5, 7)

Find the slope of

f (x) = x^{2} + 24 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 24}

Compute the slope of

f (x) = x^{2} + 24

x = 5 : m = \frac{5}{7}

Find the line with slope m=

\frac{5}{7}

and passing through

(5, 7) : y = \frac{5}{7} x + \frac{24}{7}

y = \frac{5}{7} x + \frac{24}{7}

x y^{^{'}} = 5 y - x, y (1) = 1

\int \frac{arccos ( x )}{x} d x

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

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

f (x) = \frac{3 x}{x ^{2} + 1}

What is the tangent of f(x)=sqrt(x^2+24),\at x=5 ?
The tangent of f(x)=sqrt(x^2+24),\at x=5 is y= 5/7 x+24/7