Answer
Verified
411.3k+ views
Hint: Simplify the expression by using the identity for the \[\sec \theta \] and \[\tan \theta \]. After that we need to apply here the product rule of differentiation when the differentiating the two function which is multiplied together,
\[\dfrac{d}{dx}\text{ }[f(x)\text{ }g(x)]\text{ }=\text{ }f(x)\text{ }\left[ \dfrac{d}{dx}\text{ }g(x) \right]\text{ }+\text{ }\left[ \dfrac{d}{dx}\text{ }f(x) \right]\text{ }g(x)\].
Then we need use the chain rule for differentiating function of function,
$\dfrac{d}{dx}f\left[ g(x) \right]=\dfrac{d}{dg(x)}f\left[ g(x) \right]\times \dfrac{d}{dx}g(x)$
By applying rules we have to apply the property of differentiation. Using these concepts we get the answer.
Complete step by step solution:
Let us first simplify the given expression.
Let's say that $y=\dfrac{\sec x+\tan x}{\sec x-\tan x}$.
Using $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, we get:
\[\Rightarrow \] $y=\dfrac{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}}$
\[\Rightarrow \] $y=\left( \dfrac{1+\sin x}{\cos x} \right)\times \left( \dfrac{\cos x}{1-\sin x} \right)$
\[\Rightarrow \] $y=\dfrac{1+\sin x}{1-\sin x}$
Which can also be written as:
\[\Rightarrow \] $y=(1+\sin x){{(1-\sin x)}^{-1}}$
Now, let us differentiate with respect to x.
Using the product rule of derivatives, we get:
\[\Rightarrow \] $\dfrac{dy}{dx}=(1+\sin x)\left[ \dfrac{d}{dx}{{(1-\sin x)}^{-1}} \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}$
Using the chain rule of derivatives, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=(1+\sin x)\left[ (-1){{(1-\sin x)}^{-2}}\dfrac{d}{dx}(1-\sin x) \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}\]
Using $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}k=0$, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=(1+\sin x)(-1){{(1-\sin x)}^{-2}}(-\cos x)+\cos x{{(1-\sin x)}^{-1}}\]
Which can be written as:
\[\Rightarrow \] \[\dfrac{dy}{dx}=\dfrac{(1+\sin x)(\cos x)}{{{(1-\sin x)}^{2}}}+\dfrac{\cos x}{(1-\sin x)}\]
Separating the common factor $\cos x$, and adding by equating the denominators, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=\cos x\left[ \dfrac{1+\sin x}{{{(1-\sin x)}^{2}}}+\dfrac{1-\sin x}{{{(1-\sin x)}^{2}}} \right]\]
\[\Rightarrow \] \[\dfrac{dy}{dx}=\dfrac{2\cos x}{{{(1-\sin x)}^{2}}}\], which is the required answer.
Note: The final answer can be modified in terms of other trigonometric functions, the resulting value being the same.
Derivatives of Trigonometric Functions:
$\dfrac{d}{dx}\sin x=\cos x$ $\dfrac{d}{dx}\cos x=-\sin x$
$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$
$\dfrac{d}{dx}\sec x=\tan x\sec x$ $\dfrac{d}{dx}\csc x=-\cot x\csc x$
\[\dfrac{d}{dx}\text{ }[f(x)\text{ }g(x)]\text{ }=\text{ }f(x)\text{ }\left[ \dfrac{d}{dx}\text{ }g(x) \right]\text{ }+\text{ }\left[ \dfrac{d}{dx}\text{ }f(x) \right]\text{ }g(x)\].
Then we need use the chain rule for differentiating function of function,
$\dfrac{d}{dx}f\left[ g(x) \right]=\dfrac{d}{dg(x)}f\left[ g(x) \right]\times \dfrac{d}{dx}g(x)$
By applying rules we have to apply the property of differentiation. Using these concepts we get the answer.
Complete step by step solution:
Let us first simplify the given expression.
Let's say that $y=\dfrac{\sec x+\tan x}{\sec x-\tan x}$.
Using $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, we get:
\[\Rightarrow \] $y=\dfrac{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}}$
\[\Rightarrow \] $y=\left( \dfrac{1+\sin x}{\cos x} \right)\times \left( \dfrac{\cos x}{1-\sin x} \right)$
\[\Rightarrow \] $y=\dfrac{1+\sin x}{1-\sin x}$
Which can also be written as:
\[\Rightarrow \] $y=(1+\sin x){{(1-\sin x)}^{-1}}$
Now, let us differentiate with respect to x.
Using the product rule of derivatives, we get:
\[\Rightarrow \] $\dfrac{dy}{dx}=(1+\sin x)\left[ \dfrac{d}{dx}{{(1-\sin x)}^{-1}} \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}$
Using the chain rule of derivatives, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=(1+\sin x)\left[ (-1){{(1-\sin x)}^{-2}}\dfrac{d}{dx}(1-\sin x) \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}\]
Using $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}k=0$, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=(1+\sin x)(-1){{(1-\sin x)}^{-2}}(-\cos x)+\cos x{{(1-\sin x)}^{-1}}\]
Which can be written as:
\[\Rightarrow \] \[\dfrac{dy}{dx}=\dfrac{(1+\sin x)(\cos x)}{{{(1-\sin x)}^{2}}}+\dfrac{\cos x}{(1-\sin x)}\]
Separating the common factor $\cos x$, and adding by equating the denominators, we get:
\[\Rightarrow \] \[\dfrac{dy}{dx}=\cos x\left[ \dfrac{1+\sin x}{{{(1-\sin x)}^{2}}}+\dfrac{1-\sin x}{{{(1-\sin x)}^{2}}} \right]\]
\[\Rightarrow \] \[\dfrac{dy}{dx}=\dfrac{2\cos x}{{{(1-\sin x)}^{2}}}\], which is the required answer.
Note: The final answer can be modified in terms of other trigonometric functions, the resulting value being the same.
Derivatives of Trigonometric Functions:
$\dfrac{d}{dx}\sin x=\cos x$ $\dfrac{d}{dx}\cos x=-\sin x$
$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$
$\dfrac{d}{dx}\sec x=\tan x\sec x$ $\dfrac{d}{dx}\csc x=-\cot x\csc x$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE