Question

If it is given, $y=x-{{x}^{2}}$, then the derivative of ${{y}^{2}}$ with respect to ${{x}^{2}}$ is
a. $2{{x}^{2}}+3x-1$
b. $2{{x}^{2}}-3x+1$
c. $2{{x}^{2}}+3x+1$
d. None of these

Answer 1

Hint: In order to find the solution of this question, we will first find ${{y}^{2}}$ and then we will consider ${{y}^{2}}=t$ and ${{x}^{2}}=u$. And then we will calculate $\dfrac{dt}{dx}$ and $\dfrac{du}{dx}$ to get the value of $\dfrac{dt}{du}$, that is $\dfrac{d{{y}^{2}}}{d{{x}^{2}}}$ by dividing them. And, hence, we will get the answer.

Complete step-by-step solution -
In this question, we have been asked to find the derivative of ${{y}^{2}}$ with respect to ${{x}^{2}}$, where $y=x-{{x}^{2}}$. Now, to solve this question, we will consider ${{y}^{2}}=t$ and ${{x}^{2}}=u$. Now, we have been given that $y=x-{{x}^{2}}$. So, we can say that ${{y}^{2}}={{\left( x-{{x}^{2}} \right)}^{2}}$. Now, we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. So, for a = x and $b={{x}^{2}}$, we will get,
$\begin{aligned} & {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2x\left( {{x}^{2}} \right) \\\ & {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2{{x}^{3}} \\\ \end{aligned}$
Now, as we have considered ${{y}^{2}}=t$ and ${{x}^{2}}=u$, we can write,
$t={{x}^{2}}+{{x}^{4}}-2{{x}^{3}}$ and $u={{x}^{2}}$
Now, we will find the derivative of both the equations with respect to x. We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. So, we get,
$\dfrac{dt}{dx}=2x+4{{x}^{3}}-6{{x}^{2}}$ and $\dfrac{du}{dx}=2x$
We can also write it as,
$\dfrac{dt}{dx}=4{{x}^{3}}-6{{x}^{2}}+2x........\left( i \right)$ and $\dfrac{du}{dx}=2x.........\left( ii \right)$
Now, we have been asked to find the derivative of ${{y}^{2}}$ with respect to ${{x}^{2}}$, that is, $\dfrac{d{{y}^{2}}}{d{{x}^{2}}}$ and we know that ${{y}^{2}}=t$ and ${{x}^{2}}=u$. Therefore, we can say that,
$\dfrac{d{{y}^{2}}}{d{{x}^{2}}}=\dfrac{dt}{du}$
Now we know that $\dfrac{dt}{du}$ can be calculated by calculating $\dfrac{\dfrac{dt}{dx}}{\dfrac{du}{dx}}$. So, we will substitute the values of $\dfrac{dt}{dx}$ and $\dfrac{du}{dx}$ from equations (i) and (ii). So, we get,
$\dfrac{dt}{du}=\dfrac{4{{x}^{3}}-6{{x}^{2}}+2x}{2x}$
And we know that 2x can be taken out as common from the numerator. So, we get,
$\dfrac{dt}{du}=\dfrac{2x\left( 2{{x}^{2}}-3x+1 \right)}{2x}$
And we know that common terms of the numerator and the denominator will get cancelled out, so we get,
$\dfrac{dt}{du}=2{{x}^{2}}-3x+1$
And we can further write it as,
$\dfrac{d{{y}^{2}}}{d{{x}^{2}}}=2{{x}^{2}}-3x+1$
Hence, we can say that for $y=x-{{x}^{2}}$, the derivative of ${{y}^{2}}$ with respect to ${{x}^{2}}$ is $2{{x}^{2}}-3x+1$. Therefore, option (b) is the correct answer.

Note: We can also solve this question by directly writing $x={{u}^{\dfrac{1}{2}}}$ in ${{y}^{2}}$ and then finding the derivative of ${{y}^{2}}$ with respect to u. That is, by putting $x={{u}^{\dfrac{1}{2}}}$ in ${{y}^{2}}={{x}^{2}}+{{x}^{4}}-2{{x}^{3}}$, we get ${{y}^{2}}=u+{{u}^{2}}-2{{u}^{\dfrac{3}{2}}}$ and further, on differentiating ${{y}^{2}}$ with respect to u, we get, $\dfrac{d{{y}^{2}}}{du}=1+2u-2\times \dfrac{3}{2}{{u}^{\dfrac{1}{2}}}\Rightarrow \dfrac{d{{y}^{2}}}{du}=2u-3{{u}^{\dfrac{1}{2}}}+1$. Now, we will put the value of u. So, we get, $\dfrac{d{{y}^{2}}}{d{{x}^{2}}}=2x-3x+1$ which is our answer.