What is the tangent of f(x)=sqrt(x^2+45),\at x=2 ?

The tangent of f(x)=sqrt(x^2+45),\at x=2 is y= 2/7 x+45/7

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tangent of f(x)=sqrt(x^2+45),\at x=2

tangent of

f (x) = x^{2} + 45, at x = 2

y = \frac{2}{7} x + \frac{45}{7}

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = x^{2} + 45 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 45}

Compute the slope of

f (x) = x^{2} + 45

x = 2 : m = \frac{2}{7}

Find the line with slope m=

\frac{2}{7}

and passing through

(2, 7) : y = \frac{2}{7} x + \frac{45}{7}

y = \frac{2}{7} x + \frac{45}{7}

\int_{0}^{\frac{π}{2}} sin^{7} (x) d x

\int \frac{x ^{3} + 14}{x ^{2} + 3 x + 2} d x

cos (2 t)

\frac{d y}{d x} = y^{2} + 1, y (4) = 0

x \to 1 lim (\frac{cos ( x )}{x})

What is the tangent of f(x)=sqrt(x^2+45),\at x=2 ?
The tangent of f(x)=sqrt(x^2+45),\at x=2 is y= 2/7 x+45/7