What is the tangent of f(x)=sqrt(x^2+40),\at x=3 ?

The tangent of f(x)=sqrt(x^2+40),\at x=3 is y= 3/7 x+40/7

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tangent of f(x)=sqrt(x^2+40),\at x=3

tangent of

f (x) = x^{2} + 40, at x = 3

y = \frac{3}{7} x + \frac{40}{7}

Solution steps

Find the tangent point:

(3, 7)

Find the slope of

f (x) = x^{2} + 40 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 40}

Compute the slope of

f (x) = x^{2} + 40

x = 3 : m = \frac{3}{7}

Find the line with slope m=

\frac{3}{7}

and passing through

(3, 7) : y = \frac{3}{7} x + \frac{40}{7}

y = \frac{3}{7} x + \frac{40}{7}

\frac{d y}{d x} + 7 (tan (7 x)) y = 2 cos (7 x)

x \to - 4 lim (8 - x)

\frac{d y}{d x}, x^{2} = y^{2}

\int \frac{6 ^{x}}{x} d x

\int \frac{3 x ^{2}}{x ^{2} - x - 12} d x

What is the tangent of f(x)=sqrt(x^2+40),\at x=3 ?
The tangent of f(x)=sqrt(x^2+40),\at x=3 is y= 3/7 x+40/7