Question

Differentiate with respect to x:

$\large \sf{ \frac{1 + tanx}{1 - tanx} }$
​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Rules of Derivatives had been used. First using the Quotient Rule we will simplify the equation and then differentiate it. Then using sum rule we will simplify it more. Finally after applying distributive property, we can get our answer.

Let's do it  !!

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★ Formula Used :-

$\\\;\boxed{\sf{\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{g(x)\:\bigg(\dfrac{d}{dx}\bigg)[f(x)]\;-\;f(x)\:\bigg(\dfrac{d}{dx}\bigg)[g(x)]}{g(x)^{2}}}}}$

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★ Solution :-

$\\\;\bf{\Longrightarrow\;\;\dfrac{1\;+\;\tan\:x}{1\;-\;\tan\:x}}$

Then, here

• f(x) = 1 + tan x
• g(x) = 1 - tan x

$\\\;\sf{\Longrightarrow\;\;\dfrac{f(x)}{g(x)}\;=\;\bf{\dfrac{1\;+\;\tan\:x}{1\;-\;\tan\:x}}}$

Now by using tue Formula of Quotient Rule of Derivative, we get,

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{g(x)\bigg(\dfrac{d}{dx}\bigg)[f(x)]\;-\;f(x)\bigg(\dfrac{d}{dx}\bigg)[g(x)]}{g(x)^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))\bigg(\dfrac{d}{dx}\bigg)[1\;+\;\tan (x)]\;-\;(1\;+\;\tan (x))\bigg(\dfrac{d}{dx}\bigg)[1\;-\;\tan (x)]}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))\bigg(\dfrac{d}{dx}\bigg)[(1\;+\;\tan (x)]\;-\;(1\;+\;\tan (x))\bigg(\dfrac{d}{dx}\bigg)[1\;-\;\tan (x)]}{(1\;-\;\tan (x))^{2}}}}}$

According to Sum Rule of Derivatives,

$\\\;\displaystyle{\sf{\odot\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1-\tan (x))\bigg(\dfrac{d}{dx}[1]+\dfrac{d}{dx}[\tan (x)]\bigg)-(1+\tan (x))\bigg(\dfrac{d}{dx}[1]\;+\;\dfrac{d}{dx}[-\tan (x)]\bigg)}{(1\;-\;\tan (x))^{2}}}}}$

Since, 1 is constant. Then,

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))\bigg(0\;+\;\dfrac{d}{dx}[\tan (x)]\bigg)\;-\;(1\;+\;\tan (x))\bigg(0\;+\;\dfrac{d}{dx}[-\tan (x)]\bigg)}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))\dfrac{d}{dx}[\tan (x)]\;-\;(1\;+\;\tan (x))\dfrac{d}{dx}[-\tan (x)]}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))\dfrac{d}{dx}[\tan (x)]\;+\;(1\;+\;\tan (x))\dfrac{d}{dx}[\tan (x)]}{(1\;-\;\tan (x))^{2}}}}}$

Now, derivative of tan (x) with respect to x is sec² (x).

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{(1\;-\;\tan (x))(\sec^{2}(x))\;+\;(1\;+\;\tan (x))(\sec^{2}(x)}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{\sec^{2}(x)\;-\;\tan (x)\sec^{2} (x) \;+\;\sec^{2}\;+\;\tan (x)\sec (x)}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{\sec^{2}(x)\;+\;0\;+\;\sec^{2}(x)}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\displaystyle{\bf{:\rightarrow\;\;\dfrac{d}{dx}\bigg[\dfrac{f(x)}{g(x)}\bigg]\;=\;\bf{\dfrac{2\sec^{2}(x)}{(1\;-\;\tan (x))^{2}}}}}$

$\\\;\underline{\boxed{\tt{Required\;Derivative\;=\;\bf{\dfrac{2\sec^{2}(x)}{(1\;-\;tan (x))^{2}}}}}}$

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★ More to know :-

$\\\;\mathtt{\leadsto\;\;\dfrac{d}{dx}[f(x)\;+\;g(x)]\;=\;\dfrac{d}{dx}f(x)\;+\;\dfrac{d}{dx}g(x)}$

This is Sum Rule of Derivatives.

$\\\;\mathtt{\leadsto\;\;\dfrac{d}{dx}[f(x)\;-\;g(x)]\;=\;\dfrac{d}{dx}f(x)\;-\;\dfrac{d}{dx}g(x)}$

This is Difference Rule of Derivatives.

$\\\;\mathtt{\leadsto\;\;\dfrac{d}{dx}[k\:.\:f(x)]\;=\;k\;.\;\dfrac{d}{dx}f(x)}$

This is Constant Multiplication Rule of Derivatives.

$\\\;\mathtt{\leadsto\;\;\dfrac{d}{dx}k\;=\;0}$

This is Constant Rule of Derivatives.

Answer 2

Answer:

Here the Concept of Coefficient of Elasticity of the material has been used. We know that the coefficient of Elasticity is same for all solids made of same material. First we will find the Stress and Strain of First cube. Then assuming the change in dimension of second Cube, we will find Stress and Strain of second cube. After this, we will equate them of Coefficient of Elasticity and find our answer.

Step-by-step explanation:

hope it helps you