In this post, we will learn how to find the derivative, \( \frac{dy}{dx} \), of the following function with respect to (w.r.t) \( x \):
\[
y = \tan^{-1}\left(\frac{5ax}{a^2 – 6x^2}\right).
\]
Solution:
Suppose \( u = \frac{5ax}{a^2 – 6x^2} \), then \( y = \tan^{-1}(u) \)
Taking the derivative of \( y \) w.r.t \( x\),
\[
\frac{d}{dx}(y) = \frac{d}{dx} \left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}
\]
Now compute \( \frac{du}{dx} \) using the quotient rule:
\[ \frac{du}{dx} = \frac{(a^2 – 6x^2) \cdot \frac{d}{dx}(5ax)\, – (5ax) \cdot \frac{d}{dx}(a^2 – 6x^2)}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{du}{dx} = \frac{(a^2 – 6x^2)(5a)\, – (5ax)(-12x)}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{du}{dx} = \frac{5a(a^2 – 6x^2) + 60ax^2}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{du}{dx} = \frac{5a(a^2 – 6x^2 + 12x^2)}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{du}{dx} = \frac{5a(a^2 + 6x^2)}{(a^2 – 6x^2)^2} \dots (1) \\ \]Now compute \( 1 + u^2 \):
\[ 1 + u^2 = 1 + \left(\frac{5ax}{a^2 – 6x^2}\right)^2 = 1 + \frac{25a^2x^2}{(a^2 – 6x^2)^2} \\ \Rightarrow 1 + u^2 = \frac{a^4 – 12a^2x^2 + 36x^4 + 25a^2x^2}{(a^2 – 6x^2)^2} \\ \Rightarrow 1 + u^2 = \frac{a^4 + 13a^2x^2 + 36x^4}{(a^2 – 6x^2)^2} \dots (2) \\ \]Now, substitute \( \frac{du}{dx} \) from (1) and \( 1 + u^2 \) from (2) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\frac{a^4 + 13a^2x^2 + 36x^4}{(a^2 – 6x^2)^2}} \cdot \frac{5a(a^2 + 6x^2)}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{dy}{dx} = \frac{(a^2 – 6x^2)^2}{a^4 + 13a^2x^2 + 36x^4} \cdot \frac{5a(a^2 + 6x^2)}{(a^2 – 6x^2)^2} \\ \Rightarrow \frac{dy}{dx} = \frac{5a(a^2 + 6x^2)}{a^4 + 13a^2x^2 + 36x^4} \\ \]Therefore, \( \frac{dy}{dx} \) for the given function is:
\[
\fbox{$ \frac{dy}{dx} = \frac{5a(a^2 + 6x^2)}{a^4 + 13a^2x^2 + 36x^4}$}
\]
Please let me know in the comments if you find any error in this solution.