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

The tangent of f(x)=sqrt(3x+40),\at x=3 is y= 3/14 x+89/14

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

tangent of

f (x) = 3 x + 40, at x = 3

y = \frac{3}{14} x + \frac{89}{14}

Solution steps

Find the tangent point:

(3, 7)

Find the slope of

f (x) = 3 x + 40 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 40}

Compute the slope of

f (x) = 3 x + 40

x = 3 : m = \frac{3}{14}

Find the line with slope m=

\frac{3}{14}

and passing through

(3, 7) : y = \frac{3}{14} x + \frac{89}{14}

y = \frac{3}{14} x + \frac{89}{14}

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y^{^{'}} = \frac{t ( t ^{2} + 1 )}{4 y ^{3}}

\frac{\partial}{\partial x} (x y - x y)

\frac{d}{d x} (e^{x} sin (x) + e^{x} cos (x))

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What is the tangent of f(x)=sqrt(3x+40),\at x=3 ?
The tangent of f(x)=sqrt(3x+40),\at x=3 is y= 3/14 x+89/14