Question

How do you find the derivative of $y = {\tan ^2}(3x)$ ?

Answer 1

This is a combination of three functions ${g^2}$ , $g = \tan$ and $3\;x$ . So in this case we will use the chain rule of differentiation. Then we will differentiate the functions term by term. We may have to use the chain rule many times whenever there is a conjugate function.

Formula Used : Chain rule of differentiation: The derivative of $f(g(x))$ is $f'(g(x)).g'(x)$ .
$\dfrac{d}{{dz}}({z^n}) = n{z^{n - 1}}$.
$\dfrac{d}{{dz}}(\tan z) = {\sec ^2}z$ .

Complete step by step answer:
We have;
$y = {\tan ^2}(3x)$
Let, $f = {g^2}$
And $g = \tan$
And $h(x) = 3x$ .
$\therefore g(h(x)) = \tan (3x)$ .
Then we can write;
$y = f(g(h(x)))$
At first, we will consider $g(h(x)) = j(x)$ .
$\therefore y = f(j(x))$
Differentiating both sides w.r.t. $x$ we will get;
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(f(j(x))$
Here we will apply the chain rule of differentiation and get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{df}}{{dx}}\dfrac{{dj}}{{dx}}$
Another form is;
$\Rightarrow \dfrac{{dy}}{{dx}} = f'(j(x)).j'(x)$
Now, we know that $\dfrac{d}{{dz}}({z^n}) = n{z^{n - 1}}$ . By applying this we will get;
$f'(j(x)) = 2\tan (3x)$
Now, similarly by applying the chain rule of differentiation on $j(x)$ we will get;
$j'(x) = g'(h(x))h'(x)$
We know that $\dfrac{d}{{dz}}(\tan z) = {\sec ^2}z$ .
$\therefore g'(h(x)) = {\sec ^2}(3x)$
And $h'(x) = \dfrac{d}{{dx}}(3x)$ .
Differentiating we get;
$\Rightarrow h'(x) = 3$
$\therefore j'(x) = ({\sec ^2}(3x)) \times 3$
Simplifying we get;
$\Rightarrow j'(x) = 3{\sec ^2}(3x)$
Put all this value together finally we get;
$\therefore \dfrac{{dy}}{{dx}} = 2 \times 3 \times \tan (3x) \times {\sec ^2}(3x)$
Simplifying this we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = 6 \times \tan (3x) \times {\sec ^2}(3x)$ .
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
And $\sec \theta = \dfrac{1}{{\cos \theta }}$ .
Applying this we will get;
$\Rightarrow \dfrac{{dy}}{{dx}} = 6\dfrac{{\sin 3x}}{{{{\cos }^3}(3x)}}$ .

Differentiating $y = {\tan ^2}(3x)$ we get $6\dfrac{{\sin 3x}}{{{{\cos }^3}(3x)}}$ .

Additional Information:
Differentiation of $\tan x$ w.r.t. $x$ is ${\sec ^2}x$ deduction:
We know that;
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Now to differentiate $\tan x$ w.r.t. $x$ we will use the formula $\dfrac{d}{{dz}}\left( {\dfrac{u}{v}} \right) = \dfrac{{\dfrac{{du}}{{dz}} \cdot v - u \cdot \dfrac{{dv}}{{dz}}}}{{{v^2}}}$ and get;
$\dfrac{d}{{dx}}(\tan x) = \dfrac{{\cos x \cdot \cos x - \sin x \cdot ( - \sin x)}}{{{{\cos }^2}x}}$
Simplifying we get;
$\Rightarrow \dfrac{d}{{dx}}(\tan x) = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}$
As we know that ${\sin ^2}x + {\cos ^2}x = 1$ we will get;
$\Rightarrow \dfrac{d}{{dx}}(\tan x) = \dfrac{1}{{{{\cos }^2}x}}$
We know that $\sec x = \dfrac{1}{{\cos x}}$ .
$\therefore \dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$ .

Note: This type of conjugate function will be easily solved by the chain rule of differentiation. But students must be careful about the differentiation of different functions and always mention w.r.t. what you are differentiating. Most of the time the variable is ignored by the students and then the error occurs.