Question

How do you find the equation of the line tangent to `y=(x^2-35)^7` at x=6?

Answer 1

Find the point of contact and the slope of the curve at that point to derive the equation of the tangent:

`y = 84x-503`

Explanation:

Let `f(x) = (x^2-35)^7`

Then `f'(x) = 2x*7(x^2-35)^6 = 14x(x^2-35)^6`

We find `f(6) = (6^2-35)^7 = (36-35)^7 = 1^7 = 1`

So the tangent touches the curve at `(6, 1)`

We find `f'(6) = 14*6*(6^2-35)^6 = 84`

So the slope `m` of the tangent line is `84`

The equation of a line can be written in point slope form as:

`y - y_1 = m(x - x_1)`

where `(x_1, y_1)` is a point through which the line passes and `m` is the slope.

So the equation of our tangent line can be written:

`y - 1 = 84(x - 6)`

Adding `1` to both sides and simplifying gives us the equation in slope intercept form:

`y = 84x - 503`