Question

Find the derivative of $y=\dfrac{{{t}^{2}}-1}{t+1}$.

Answer 1

Hint: To find the derivative of the given function, we have to use the formula, $d\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{g\left( x \right).df\left( x \right)-f\left( x \right).dg\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$. Here, we have $f\left( x \right)$ as ${{t}^{2}}-1$ and $g\left( x \right)$ as $t+1$. So, we will substitute them in the formula and find the derivative. We must also know that derivative of ${{x}^{n}}=n{{x}^{n-1}}$.

Complete step-by-step solution:
We have been asked to find the derivative of the function given in the question. We have been given that the function is $y=\dfrac{{{t}^{2}}-1}{t+1}$. Here, y is a function of t. Now, we can see that the function is in the from of a fraction. The numerator is given as ${{t}^{2}}-1$ and the denominator is given as $t+1$. So, now to differentiate such form of functions, we have a rule called the quotient rule. Let us first understand the rule. Let us suppose two functions in x as $f\left( x \right)$ and $g\left( x \right)$, then the derivative of $\dfrac{f\left( x \right)}{g\left( x \right)}$ is given as,
 $\dfrac{dy}{dt}=d\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{g\left( x \right).df\left( x \right)-f\left( x \right).dg\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}\ldots \ldots \ldots \left( i \right)$
So, now here we have, $f\left( t \right)={{t}^{2}}-1\Rightarrow df\left( t \right)=2t$ and $g\left( t \right)=t+1\Rightarrow dg\left( t \right)=t$. So, we will now substitute these values in the equation (i). So, we will get,
$\dfrac{dy}{dt}=\dfrac{\left( t+1 \right)2t-\left( {{t}^{2}}-1 \right).1}{{{\left( t+1 \right)}^{2}}}$
Simplifying this further we will get as,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{2{{t}^{2}}+2t-{{t}^{2}}+1}{{{\left( t+1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{t}^{2}}+2t+1}{{{\left( t+1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{\left( t+1 \right)}^{2}}}{{{\left( t+1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=1 \\
\end{align}$
Therefore, we get the derivative of the given function, $y=\dfrac{{{t}^{2}}-1}{t+1}$ as 1.

Note: We have used differentiated with respect to ‘t’ here in this question since the function was defined on the variable t, if not we should use the suitable variable to differentiate the function accordingly. Most of the students go wrong in applying the quotient rule. They tend to change the signs of the terms in the numerator. Instead of a negative signs, they use the positive sign which makes the whole solution incorrect. So, students should be careful and avoid this mistake.