Question

If \[y = \dfrac{{{e^{2x}} + {e^{ - 2x}}}}{{{e^{2x}} - {e^{ - 2x}}}}\], then \[\dfrac{{dy}}{{dx}} = \]
a) \[\dfrac{{ - 8}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
b) \[\dfrac{8}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
c) \[\dfrac{{ - 4}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
d) \[\dfrac{4}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]

Answer 1

Hint: Here the question is related to differentiation, we have to determine the derivative of the given function. The given term is in the form of fraction firstly we apply the quotient rule and then applying the differentiation formulas to the exponential function then on simplification we get a final result.

Complete step-by-step answer:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
An exponential function is a Mathematical function in the form \[f(x) = a\] where “x” is a variable and “a” is a constant which is called the base of the function and it should be greater than 0. The most commonly used exponential function base is the transcendental number e, which is approximately equal to 2.71828.
On considering the given question, we have
\[ \Rightarrow y = \dfrac{{{e^{2x}} + {e^{ - 2x}}}}{{{e^{2x}} - {e^{ - 2x}}}}\]
On differentiating the function y with respect to x. Here the function is in the form of fraction so initially we use the quotient rule.
The quotient rule is given by \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\dfrac{{du}}{{dx}} - u.\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{({e^{2x}} - {e^{ - 2x}}).\dfrac{d}{{dx}}({e^{2x}} + {e^{ - 2x}}) - ({e^{2x}} + {e^{ - 2x}}).\dfrac{d}{{dx}}({e^{2x}} - {e^{ - 2x}})}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
In the numerator the derivative is in the form of \[\dfrac{d}{{dx}}(u \pm v)\] and it is given by \[\dfrac{d}{{dx}}(u \pm v) = \dfrac{{du}}{{dx}} \pm \dfrac{{dv}}{{dx}}\]. So we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{({e^{2x}} - {e^{ - 2x}}).\dfrac{d}{{dx}}({e^{2x}}) + \dfrac{d}{{dx}}({e^{ - 2x}}) - ({e^{2x}} + {e^{ - 2x}}).\dfrac{d}{{dx}}({e^{2x}}) - \dfrac{d}{{dx}}({e^{ - 2x}})}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
By the standard formula for the exponential function is given by \[\dfrac{d}{{dx}}({e^{ax}}) = a{e^{ax}}\], on using this formula we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{({e^{2x}} - {e^{ - 2x}}).\left[ {2({e^{2x}}) + ( - 2)({e^{ - 2x}})} \right] - ({e^{2x}} + {e^{ - 2x}}).\left[ {2({e^{2x}}) - ( - 2)({e^{ - 2x}})} \right]}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2[({e^{2x}} - {e^{ - 2x}}).({e^{2x}} - {e^{ - 2x}}) - ({e^{2x}} + {e^{ - 2x}}).({e^{2x}} + {e^{ - 2x}})]}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
The term is multiplied twice so it is written in the form of square
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2[{{({e^{2x}} - {e^{ - 2x}})}^2} - {{({e^{2x}} + {e^{ - 2x}})}^2}]}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
The term is in the form of \[({a^2} - {b^2})\]. On using the formula \[{a^2} - {b^2} = (a + b)(a - b)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2[({e^{2x}} - {e^{ - 2x}} + {e^{2x}} + {e^{ - 2x}})({e^{2x}} - {e^{ - 2x}} - {e^{2x}} - {e^{ - 2x}})]}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
On cancelling the terms and simplifying
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2[(2{e^{2x}})( - 2{e^{ - 2x}})]}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 8}}{{{{({e^{2x}} - {e^{ - 2x}})}^2}}}\]
Hence the option a) is the correct one.
So, the correct answer is “Option a”.

Note: The student must know about the differentiation formulas for the power, trigonometric and exponential function and these differentiation formulas are standard. If the function is in the form of product and the fraction we use the product rule: \[\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] and the quotient rule : \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\dfrac{{du}}{{dx}} - u.\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]