# What is the equation of the tangent line of #f(x)=ln(x+3)/x+7x# at #x=3#?

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Equation of the tangent line passing through the point

Multiplying the functin with x

Differentiating wrt x on both sides

Substituting for x and y

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To find the equation of the tangent line of f(x) = ln(x+3)/(x+7x) at x=3, we need to find the derivative of f(x) and evaluate it at x=3.

First, let's find the derivative of f(x) using the quotient rule:

f'(x) = [(x+7x)(1/(x+3)) - ln(x+3)(1+7)] / (x+7x)^2

Simplifying this expression, we get:

f'(x) = [(8x+1)/(x+3)^2] - [ln(x+3)(8)] / (x+7x)^2

Now, let's evaluate f'(x) at x=3:

f'(3) = [(8(3)+1)/(3+3)^2] - [ln(3+3)(8)] / (3+7(3))^2

Simplifying this expression, we get:

f'(3) = [25/36] - [ln(6)(8)] / 169

Therefore, the slope of the tangent line at x=3 is f'(3) = [25/36] - [ln(6)(8)] / 169.

To find the equation of the tangent line, we use the point-slope form:

y - f(3) = f'(3)(x - 3)

Now, substitute the values of f(3) and f'(3) into the equation:

y - [ln(3+3)/(3+7(3))] = [25/36] - [ln(6)(8)] / 169 * (x - 3)

Simplifying this equation, we get the equation of the tangent line of f(x) at x=3.

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