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High School
Calculus and Analysis
Calculus 1
Derivatives

Leia Sullivan

Answered question

2022-04-13

How do you find the equation of the line tangent to

y = {(x^{2} - 35)}^{7}

at x=6?

Answer & Explanation

analiticozuod

Beginner2022-04-14Added 10 answers

Let

f (x) = {(x^{2} - 35)}^{7}

Then

f^{'} (x) = 2 x \cdot 7 {(x^{2} - 35)}^{6} = 14 x {(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 \cdot 6 \cdot {(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

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