To find the derivative
‌ when
y=tan−1(‌), we need to use the chain rule and the formula for the derivative of the inverse tangent function.
First, recall the formula for the derivative of the inverse tangent function:
‌(tan−1(u))=‌⋅‌ Here, our
u=‌. We'll need to use the quotient rule to find
‌.
The quotient rule states:
‌(‌)=‌| g(x)f′(x)−f(x)g′(x) |
| (g(x))2 |
where
f(x)=3−2x and
g(x)=1+6x. Therefore,
f′(x)=−2and
g′(x)=6‌. ‌Now apply the quotient rule:
‌=‌| (1+6x)(−2)−(3−2x)(6) |
| (1+6x)2 |
Simplify the numerator:
‌=‌| −2−12x−18+12x |
| (1+6x)2 |
=‌Now, substituting
u=‌ into the derivative formula for the inverse tangent:
‌=‌⋅‌ Simplify the term inside the inverse tangent formula:
(‌)2=‌=‌So the expression for the derivative becomes:
‌=‌⋅‌Combine the terms in the denominator:
‌=‌| 1 |
| (1+6x)2+9−12x+4x2 | | (1+6x)2 |
|
⋅‌ Evaluate the combined fraction:
(1+6x)2+9−12x+4x2=1+12x+36x2+9−12x+4x2=10+40x2
Thus the expression simplifies to:
The
(1+6x)2 terms cancel out:
‌=‌=−‌Therefore, the correct answer is:
Option C
−‌