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How do you find the derivative of \[y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}\]?

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Last updated date: 24th Jul 2024
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Answer
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Hint: To solve this problem, we should know the derivatives of some of the functions. We should also know the quotient rule of differentiation which is used to differentiate expressions of form \[\dfrac{f(x)}{g(x)}\]. The function whose derivatives we should know is \[{{x}^{2}}\], its derivative with respect to x is \[2x\]. The quotient rule states that the expressions of the form \[\dfrac{f(x)}{g(x)}\] are differentiated as,
\[\dfrac{d\left( \dfrac{f(x)}{g(x)} \right)}{dx}=\dfrac{f'\left( x \right)g\left( x \right)-g'\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}\].

Complete step by step solution:
We are given the expression \[y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}\], we need to find its derivative. This expression is of the form \[\dfrac{f(x)}{g(x)}\]. Here, \[f(x)={{x}^{2}}-1\And g(x)={{x}^{2}}+1\]. We will use the quotient rule to differentiate this expression. We know that the quotient rule states that expressions of the form \[\dfrac{f(x)}{g(x)}\] are differentiated as, \[\dfrac{d\left( \dfrac{f(x)}{g(x)} \right)}{dx}=\dfrac{f'\left( x \right)g\left( x \right)-g'\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}\].
We know that the derivative of \[{{x}^{2}}\] with respect to x is \[2x\]. Thus, the derivative of \[{{x}^{2}}-1\] with respect to x is also \[2x\]. Similarly, the derivative of \[{{x}^{2}}+1\] with respect to x is also \[2x\].
Using the quotient rule on the expression \[y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}\], we get
\[\dfrac{d\left( y \right)}{dx}=\dfrac{\dfrac{d\left( {{x}^{2}}-1 \right)}{dx}\left( {{x}^{2}}+1 \right)-\dfrac{d\left( {{x}^{2}}+1 \right)}{dx}\left( {{x}^{2}}-1 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Substituting the derivatives of the functions, we get
\[\dfrac{dy}{dx}=\dfrac{2x\left( {{x}^{2}}+1 \right)-2x\left( {{x}^{2}}-1 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Expanding the brackets, we get
\[\dfrac{dy}{dx}=\dfrac{2{{x}^{3}}+2x-2{{x}^{3}}+2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Simplifying the above expression, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{4x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Thus, the derivative of the expression \[y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}\] is \[\dfrac{dy}{dx}=\dfrac{4x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\].

Note:
To solve these questions, one should know the derivatives of the functions and quotient rules. We can also solve the differentiate the expression as,
\[y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}\]
Adding and subtracting 1 in the numerator, we get
\[\Rightarrow y=\dfrac{\left( {{x}^{2}}+1 \right)+\left( -1-1 \right)}{{{x}^{2}}+1}\]
\[\Rightarrow y=\dfrac{{{x}^{2}}+1}{{{x}^{2}}+1}-\dfrac{2}{{{x}^{2}}+1}\]
\[\Rightarrow y=1-\dfrac{2}{{{x}^{2}}+1}\]
Differentiating both sides of the above expression, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d(1)}{dx}-\dfrac{d\left( \dfrac{2}{{{x}^{2}}+1} \right)}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}=0+\dfrac{2\times 2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=\dfrac{4x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Thus, we are getting the same answer from both methods.