Question

What is the derivative of $y = \arccos {(x)^2}?$

Answer 1

Hint: As we can see that we have to solve the given integral. We can solve this integral by using the formula of integration by parts and doing some calculations we will get the required answer. We will be using the chain rule formula. It says that $[f(g(x))] = f'(g(x)) \cdot g'(x)$. So we have to calculate derivative, so we can write it as $\dfrac{d}{{dx}}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Complete step by step solution:
Here we have to find the integral of $y = \arccos {(x)^2}$.
Here we have $f(x) = \arccos (x)$ and $g(x) = {x^2}$.
Let us assume that $u = {x^2}$, now by applying chain rule we can write $\dfrac{d}{{du}}[\arccos (u)]\dfrac{d}{{dx}}[{x^2}]$.
We know that the derivative of $\arccos (u)$ with respect to $u$ is $ - \dfrac{1}{{\sqrt {1 - {u^2}} }}$. So by putting this in the formula we can write $ - \dfrac{1}{{\sqrt {1 - {u^2}} }}\dfrac{d}{{dx}}[{x^2}]$.
Now we replace $u = {x^2}$, so we have $ - \dfrac{1}{{\sqrt {1 - {{({x^2})}^2}} }}\dfrac{d}{{dx}}[{x^2}]$. ON further solving we can write $ - \dfrac{1}{{\sqrt {1 - ({x^4})} }}\dfrac{d}{{dx}}[{x^2}]$.
By differentiating with power rules we know that $\dfrac{d}{{dx}}[{x^n}]$ is written as $n{x^{n - 1}}$. By comparing here we have $n = 2$.
So we can write it as $ - \dfrac{1}{{\sqrt {1 - {x^4}} }}2{x^{2 - 1}} = - \dfrac{1}{{\sqrt {1 - {x^4}} }}2{x^1}$.
Hence the required answer is $ - \dfrac{{2x}}{{\sqrt {1 - {x^4}} }}$.

Note:
We should note the exponential formula we used above i.e. ${({x^2})^2}$ can be written as ${x^{2 \times 2}} = {x^4}$. Before solving this kind of question we should be fully aware of the integration and their formulas. We should avoid the calculation mistake. All the basic integration and derivative formulas should be memorized to solve these types of questions.