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

How do you find the derivative of y=x2−1x2+1y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}?

Explanation

Solution

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 f(x)g(x)\dfrac{f(x)}{g(x)}. The function whose derivatives we should know is x2{{x}^{2}}, its derivative with respect to x is 2x2x. The quotient rule states that the expressions of the form f(x)g(x)\dfrac{f(x)}{g(x)} are differentiated as,
d(f(x)g(x))dx=f′(x)g(x)−g′(x)f(x)(g(x))2\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=x2−1x2+1y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}, we need to find its derivative. This expression is of the form f(x)g(x)\dfrac{f(x)}{g(x)}. Here, f(x)=x2−1&g(x)=x2+1f(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 f(x)g(x)\dfrac{f(x)}{g(x)} are differentiated as, d(f(x)g(x))dx=f′(x)g(x)−g′(x)f(x)(g(x))2\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 x2{{x}^{2}} with respect to x is 2x2x. Thus, the derivative of x2−1{{x}^{2}}-1 with respect to x is also 2x2x. Similarly, the derivative of x2+1{{x}^{2}}+1 with respect to x is also 2x2x.
Using the quotient rule on the expression y=x2−1x2+1y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}, we get
d(y)dx=d(x2−1)dx(x2+1)−d(x2+1)dx(x2−1)(x2+1)2\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
dydx=2x(x2+1)−2x(x2−1)(x2+1)2\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
dydx=2x3+2x−2x3+2x(x2+1)2\dfrac{dy}{dx}=\dfrac{2{{x}^{3}}+2x-2{{x}^{3}}+2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}
Simplifying the above expression, we get
⇒dydx=4x(x2+1)2\Rightarrow \dfrac{dy}{dx}=\dfrac{4x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}
Thus, the derivative of the expression y=x2−1x2+1y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} is dydx=4x(x2+1)2\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=x2−1x2+1y=\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1}
Adding and subtracting 1 in the numerator, we get
⇒y=(x2+1)+(−1−1)x2+1\Rightarrow y=\dfrac{\left( {{x}^{2}}+1 \right)+\left( -1-1 \right)}{{{x}^{2}}+1}
⇒y=x2+1x2+1−2x2+1\Rightarrow y=\dfrac{{{x}^{2}}+1}{{{x}^{2}}+1}-\dfrac{2}{{{x}^{2}}+1}
⇒y=1−2x2+1\Rightarrow y=1-\dfrac{2}{{{x}^{2}}+1}
Differentiating both sides of the above expression, we get
⇒dydx=d(1)dx−d(2x2+1)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d(1)}{dx}-\dfrac{d\left( \dfrac{2}{{{x}^{2}}+1} \right)}{dx}
⇒dydx=0+2×2x(x2+1)2=4x(x2+1)2\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.