Answer
Verified
406.8k+ views
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,
$\begin{align}
& {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2x\left( {{x}^{2}} \right) \\
& {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2{{x}^{3}} \\
\end{align}$
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.
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,
$\begin{align}
& {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2x\left( {{x}^{2}} \right) \\
& {{y}^{2}}={{x}^{2}}+{{x}^{4}}-2{{x}^{3}} \\
\end{align}$
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE