Question

Find $\dfrac{d}{d x}\left(x^{3} \tan ^{2} \dfrac{x}{2}\right)=$.
(1) ${{x}^{3}}\tan \left( \dfrac{x}{2} \right){{\sec }^{2}}\left( \dfrac{x}{2} \right)+3x{{\tan }^{2}}\left( \dfrac{x}{2} \right)$
(2) ${{x}^{3}}{{\tan }^{2}}\left( \dfrac{x}{2} \right){{\sec }^{2}}\left( \dfrac{x}{2} \right)+3{{x}^{2}}{{\tan }^{2}}\left( \dfrac{x}{2} \right)$
(3) ${{x}^{3}}\tan \left( \dfrac{x}{2} \right){{\sec }^{2}}\left( \dfrac{x}{2} \right)+3{{x}^{2}}{{\tan }^{2}}\left( \dfrac{x}{2} \right)$
(4) none of these

Answer 1

Hint: We can solve this using the product rule. That is $\dfrac{d}{d x} f(x) \cdot g(x)=\left[g(x) \times f^{\prime}(x)+f(x) \times g^{\prime}(x)\right]$. Here \[f(x)={{x}^{3}}\] and \[g(x)={{\tan }^{2}}\left( \dfrac{x}{2} \right)\]. After applying this rule and then by using the chain rule extended formula, we can find the solution for the given question.

Formula used: $\dfrac{{d(f(x).g(x))}}{{dx}} = g(x) \times {f’}(x) + f(x) \times {g’}(x)$

Complete Step by step solution:
Let $y={{x}^{3}}{{\tan }^{2}}\left( \dfrac{x}{2} \right)$
Differentiate with respect to x.
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{3}}{{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)$
Applying the product rule we have,
\[\dfrac{dy}{dx}=\dfrac{d{{x}^{3}}}{dx}.{{\tan }^{2}}\left( \dfrac{x}{2} \right)+\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right).{{x}^{3}}\]
We know that the differentiation of \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\]. Applying this we have,
\[\dfrac{dy}{dx}=3{{x}^{2}}.{{\tan }^{2}}\left( \dfrac{x}{2} \right)+\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right).{{x}^{3}}\,\,\,\,--(1)\].
But \[\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)\] can be solved by chain rule extended formula.
That is $\dfrac{d}{d x}[f(g(h(x)))]=f^{\prime}(g(h(x))) g^{\prime}(h(x)) h^{\prime}(x)$, then we have
\[\Rightarrow \dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)=2\tan \left( \dfrac{x}{2} \right).\dfrac{d}{dx}\left( \tan \left( \dfrac{x}{2} \right) \right)\]
We know that differentiation of \[\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\], then we have
\[\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)=2\tan \left( \dfrac{x}{2} \right).{{\sec }^{2}}\left( \dfrac{x}{2} \right)\dfrac{d}{dx}\left( \dfrac{x}{2} \right)\]
\[\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)=2\tan \left( \dfrac{x}{2} \right).{{\sec }^{2}}\left( \dfrac{x}{2} \right).\dfrac{1}{2}\]
\[\dfrac{d}{dx}\left( {{\tan }^{2}}\left( \dfrac{x}{2} \right) \right)=\tan \left( \dfrac{x}{2} \right).{{\sec }^{2}}\left( \dfrac{x}{2} \right)\]
Substituting this in equation 1 we have,
\[\Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}.{{\tan }^{2}}\left( \dfrac{x}{2} \right)+\tan \left( \dfrac{x}{2} \right).{{\sec }^{2}}\left( \dfrac{x}{2} \right).{{x}^{3}}\]
\[\Rightarrow \dfrac{dy}{dx}={{x}^{3}}{{\tan }}\left( \dfrac{x}{2} \right){{\sec }^{2}}\left( \dfrac{x}{2} \right)+3{{x}^{2}}{{\tan }^{2}}\left( \dfrac{x}{2} \right)\]

Hence, option (3) is correct.

Note: We can differentiate between two or more functions in a given function using product rules. For the sum of the two functions, the formula for the product rule appears to be as follows. The same pattern applies if the three functions have produced a product. So, we need to apply the extended chain rule. Also, note that the differentiation of any constant is zero.