Solveeit Logo
Login

Question

Question: If \[y=\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\dfrac{dy}{dx}=\] A) \[\dfrac{{{e}^{x}}}...

If y=1+ex1−exy=\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}}, then dydx=\dfrac{dy}{dx}=
A) ex(1−ex)(1−e2x)\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{(1-{{e}^{2x}})}}
B) ex(1−ex)(1−ex)\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{(1-{{e}^{x}})}}
C) ex(1−ex)(1+e2x)\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{(1+{{e}^{2x}})}}
D) ex(1−ex)(1+ex)\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{(1+{{e}^{x}})}}

Explanation

Solution

According to the given question, here we have given the value of the derivative. Before taking that first of all we need to square on both side then the equation becomes easier and apply the derivation rule that is uv\dfrac{u}{v} rule and solve further you will get the value of dydx\dfrac{dy}{dx}.

Complete step by step answer:
According to the given question that is y=1+ex1−exy=\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}}
First of all in this particular problem we have to remove the root sign to avoid complexity
We have to square on both sides on this equation we get:
y2=(1+ex1−ex)2{{y}^{2}}={{\left( \sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}} \right)}^{2}}
By simplifying this and solving we get:
y2=(1+ex1−ex){{y}^{2}}=\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)
Now, differentiate with respect to x we get:
d(y2)dx=d(1+ex1−ex)dx−−(1)\dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}--(1)
After simplifying and also apply derivation rule that is d(y2)dx=2ydydx\dfrac{d\left( {{y}^{2}} \right)}{dx}=2y\dfrac{dy}{dx} on LHS and also apply uv\dfrac{u}{v}rule on RHS
On LHS d(y2)dx=2ydydx−−(2)\dfrac{d\left( {{y}^{2}} \right)}{dx}=2y\dfrac{dy}{dx}--(2)
uv\dfrac{u}{v}Rule means d(uv)dx=vd(u)dx−ud(v)dxv2\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\dfrac{d(u)}{dx}-u\dfrac{d(v)}{dx}}{{{v}^{2}}}
By applying on RHS we get:
d(1+ex1−ex)dx=(1−ex)d(1+ex)dx−(1+ex)d(1−ex)dx(1−ex)2\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{(1-{{e}^{x}})\dfrac{d(1+{{e}^{x}})}{dx}-(1+{{e}^{x}})\dfrac{d(1-{{e}^{x}})}{dx}}{{{(1-{{e}^{x}})}^{2}}}
After taking the derivative and simplifying we get:
d(1+ex1−ex)dx=(1−ex)d(ex)dx−(1+ex)(−d(ex)dx)(1−ex)2\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{(1-{{e}^{x}})\dfrac{d({{e}^{x}})}{dx}-(1+{{e}^{x}})\left( -\dfrac{d({{e}^{x}})}{dx} \right)}{{{(1-{{e}^{x}})}^{2}}}
But d(ex)dx=ex\dfrac{d({{e}^{x}})}{dx}={{e}^{x}}therefore apply in this above equation we get:
d(1+ex1−ex)dx=(1−ex)(ex)−(1+ex)(−ex)(1−ex)2\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{(1-{{e}^{x}})({{e}^{x}})-(1+{{e}^{x}})\left( -{{e}^{x}} \right)}{{{(1-{{e}^{x}})}^{2}}}
After further solving this we get:
d(1+ex1−ex)dx=(1−ex)(ex)+(1+ex)(ex)(1−ex)2\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{(1-{{e}^{x}})({{e}^{x}})+(1+{{e}^{x}})\left( {{e}^{x}} \right)}{{{(1-{{e}^{x}})}^{2}}}
Here, take ex{{e}^{x}}common we get:
d(1+ex1−ex)dx=(ex)((1−ex)+(1+ex))(1−ex)2\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{({{e}^{x}})((1-{{e}^{x}})+(1+{{e}^{x}}))}{{{(1-{{e}^{x}})}^{2}}}
After simplifying further we get:
d(1+ex1−ex)dx=2ex(1−ex)2−−(3)\dfrac{d\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{dx}=\dfrac{2{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}--(3)
Substitute the equation (2) and equation (3) on equation (1)
2ydydx=2ex(1−ex)22y\dfrac{dy}{dx}=\dfrac{2{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}
Here 2 get cancelled and simplifying we get:
dydx=ex(y)(1−ex)2\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{(y){{(1-{{e}^{x}})}^{2}}}
Substitute the value of y in this above equation we get:
dydx=ex(1+ex1−ex)(1−ex)2\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{\sqrt{\left( \dfrac{1+{{e}^{x}}}{1-{{e}^{x}}} \right)}{{(1-{{e}^{x}})}^{2}}}
dydx=ex(1−ex)2(1−ex1+ex)\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}\sqrt{\left( \dfrac{1-{{e}^{x}}}{1+{{e}^{x}}} \right)}
If you go through the option then it doesn’t match then we have to simplify further this above equation we get:
That means we have to rationalize the denominator to get the desired answer.
dydx=ex(1−ex)2(1−ex1+ex)×(1−ex1−ex)\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}\sqrt{\left( \dfrac{1-{{e}^{x}}}{1+{{e}^{x}}} \right)\times \left( \dfrac{1-{{e}^{x}}}{1-{{e}^{x}}} \right)}
After simplifying
dydx=ex(1−ex)2(1−ex)2(1+ex)(1−ex)\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}\sqrt{\dfrac{{{\left( 1-{{e}^{x}} \right)}^{2}}}{\left( 1+{{e}^{x}} \right)\left( 1-{{e}^{x}} \right)}}
By applying the property and further simplifying we get:
(a−b)(a+b)=a2−b2(a-b)(a+b)={{a}^{2}}-{{b}^{2}}
dydx=ex(1−ex)(1+ex)(1−ex)\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{\left( 1+{{e}^{x}} \right)\left( 1-{{e}^{x}} \right)}}
Therefore, we get:
dydx=ex(1−ex)((1)2−(ex)2)\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{\left( {{(1)}^{2}}-{{({{e}^{x}})}^{2}} \right)}}
Therefore, final answer is
∴dydx=ex(1−ex)(1−e2x)\therefore \dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{\left( 1-{{e}^{2x}} \right)}}
Therefore, the correct option is option (A).

Note:
In this particular problem the equation is given. It is more advisable that if the roots are present then we have to square on both sides then the problem becomes simpler. Hence, after squaring then taking the derivative is more preferable. Another method to do the same problem is that you can take a direct derivative on RHS with roots no need of squaring but then it will become complex . To make the sum easier we need to square then derivative and also apply the formula like uv\dfrac{u}{v} rule and don’t make silly mistake while simplify this uv\dfrac{u}{v}rule. So, the above solution is preferable for such types of problems.