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Question: How do you find the derivative of \(\left( \dfrac{{{t}^{2}}+1}{{{t}^{2}}-1} \right)?\)...

How do you find the derivative of (t2+1t2−1)?\left( \dfrac{{{t}^{2}}+1}{{{t}^{2}}-1} \right)?

Explanation

Solution

(1)\left( 1 \right) By using the quotient rule we have
(u(x)v(x))′=u′(x).v(x)−u(x)v′(x)(v(x))2{{\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)}^{'}}=\dfrac{u'\left( x \right).v\left( x \right)-u\left( x \right)v'\left( x \right)}{{{\left( v\left( x \right) \right)}^{2}}}
Where u(x)u\left( x \right) and v(x)v\left( x \right) are functions and u′(x),v′(x)u'\left( x \right),v'\left( x \right) are its respective derivatives.
(2)\left( 2 \right) We have to apply the quotient rule when one function is divided by another function, then we have to take derivation of that.

Complete step-by-step solution:
Here, we have
(t2+1t2−1)\left( \dfrac{{{t}^{2}}+1}{{{t}^{2}}-1} \right)
Equation can be written as,
∴f(12)=(t2+1t2−1)\therefore f\left( \dfrac{1}{2} \right)=\left( \dfrac{{{t}^{2}}+1}{{{t}^{2}}-1} \right)
Here we can see that one function divides by another.
Now, we have to use quotient rules.
The quotient rule we have,
(u(x)v(x))′=u′(x).v(x)−u(x)v′(x)(v(x))2{{\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)}^{'}}=\dfrac{u'\left( x \right).v\left( x \right)-u\left( x \right)v'\left( x \right)}{{{\left( v\left( x \right) \right)}^{2}}}
Where u(x)u\left( x \right) and v(x)v\left( x \right) are functions and u′(x),v′(x)u'\left( x \right),v'\left( x \right) are its respective derivatives.
Now, we can apply it in given equation we get,
f′(t)=(t2−1)(2t)−(t2+1)(2t)(t2−1)2f'\left( t \right)=\dfrac{\left( {{t}^{2}}-1 \right)\left( 2t \right)-\left( {{t}^{2}}+1 \right)\left( 2t \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}
∴f′(t)=2t(t2−1−t2−1)(t2−1)2\therefore f'\left( t \right)=\dfrac{2t\left( {{t}^{2}}-1-{{t}^{2}}-1 \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}
∴f′(t)=−4t(t2−1)2\therefore f'\left( t \right)=\dfrac{-4t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}

The derivative of (t2+1t2−1)\left( \dfrac{{{t}^{2}}+1}{{{t}^{2}}-1} \right) is −4t(t2−1)2\dfrac{-4t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}

Note: Differentiations is a process in which we divide a big number or a big value into small numbers of values, i.e. diving the bigger values in parts in form of small parts is called as differentiation
While integration is exactly opposite to the differentiation, integration means to integrate things, or to add things, in integration we add all small parts and values to form a bigger-parts.
Hence, integration and differentiation are exactly opposite to each other, and the properties of integration and differentiations are exactly opposite to each other.
(1)\left( 1 \right) The sum derivative says that it is a contract between two or more parties in which the value is based on financial assets.
(2)\left( 2 \right) Differentiation and integration consists of the two operations in a single variable.