Tangent Line Help

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Find the equation of the tangent line to the graph of f(x) = ?6x ln(x + 5) at x = ?2.

 

[mmm4444bot]

Again, function f is defined as a product.

f(x) = h(x) * g(x)

where

h(x) = -6x

g(x) = ln(x + 5)

The slope of tangent lines to f is f`(x).

So, you need to find the derivative of f, in order to calculate the slope when x = 2.

You can evaluate f(2) right now. That's the y-coordinate at the tangent point. x = 2 is the other coordinate, obviously.

Use the value of f`(2) as m, and use x1 = 2 and y1 = f(2) in the Point-Slope formula:

y - y1 = m(x - x1) to get the linear equation.

The Product Rule says:

f`(x) = h(x) * g`(x) + h`(x) * g(x)

The derivative of ln(x) is 1/x. Shifting x by five units does not change the shape of the curve.

I mean, the derivative of ln(x + c) = 1/(x + c), where c is a constant.

And, when we apply the Chain Rule, because ln(x + 5) is a composite function, the derivative of the inner function is simply 1, so we don't bother multiplying by it.

You know the derivative of h(x), yes?

Can you put it all together?