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Hint: To find the derivative of the given function, we will simplify the given function in terms of fractions using trigonometric relations and then find the derivative using quotient rule of differentiation.

We have the function \[y=\sec x\]. We have to find the first derivative of the given function.

Thus, we will differentiate the given function with respect to the variable \[x\].

We can rewrite \[y=\sec x\] in terms of \[\cos x\] as \[y=\sec x=\dfrac{1}{\cos x}\].

We will now use quotient rule to find the derivative of the given function which states that if \[y=\dfrac{f\left( x \right)}{g\left( x \right)}\], then we have \[\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}\].

We have to evaluate \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{dy}{dx}\left( \dfrac{1}{\cos x} \right)\].

Thus, substituting \[f\left( x \right)=1,g\left( x \right)=\cos x\] in the quotient rule of differentiation, we get \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{dy}{dx}\left( \dfrac{1}{\cos x} \right)=\dfrac{\cos x\times \dfrac{d}{dx}\left( 1 \right)-1\times \dfrac{d}{dx}\left( \cos x \right)}{{{\left( \cos x \right)}^{2}}}\]. \[...\left( 1 \right)\]

We know that differentiation of a constant is zero with respect to any variable. Thus, we have\[\dfrac{d}{dx}\left( 1 \right)=0\]. \[...\left( 2 \right)\]

We also know that differentiation of function of the form \[y=\cos x\] is \[\dfrac{dy}{dx}=-\sin x\]. \[...\left( 3 \right)\]

Substituting the value of equation \[\left( 2 \right), \left( 3 \right)\] in equation \[\left( 1 \right)\], we have \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{\cos x\times \dfrac{d}{dx}\left( 1 \right)-1\times \dfrac{d}{dx}\left( \cos x \right)}{{{\left( \cos x \right)}^{2}}}=\dfrac{\cos x\times 0-1\times \left( -\sin x \right)}{{{\cos }^{2}}x}=\dfrac{\sin x}{{{\cos }^{2}}x}\].

We know that \[\dfrac{\sin x}{\cos x}=\tan x\]. Thus, we have \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{\sin x}{{{\cos }^{2}}x}=\dfrac{\tan x}{\cos x}=\tan x\sec x\].

Hence, the derivative of the function \[y=\sec x\] is \[\dfrac{dy}{dx}\left( \sec x \right)=\tan x\sec x\].

The derivative of any function \[y=f\left( x \right)\] with respect to variable \[x\] is a measure of the rate at which the value of the function changes with respect to the change in the value of variable \[x\]. The first derivative of any function also signifies the slope of the function when the graph of \[y=f\left( x \right)\] is plotted against \[x\] considering only real values of the function.

Note: It’s necessary to use quotient rules to find the derivative of the given function. We can also use the basic formula for finding the derivative of any function using limit.

We have the function \[y=\sec x\]. We have to find the first derivative of the given function.

Thus, we will differentiate the given function with respect to the variable \[x\].

We can rewrite \[y=\sec x\] in terms of \[\cos x\] as \[y=\sec x=\dfrac{1}{\cos x}\].

We will now use quotient rule to find the derivative of the given function which states that if \[y=\dfrac{f\left( x \right)}{g\left( x \right)}\], then we have \[\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}\].

We have to evaluate \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{dy}{dx}\left( \dfrac{1}{\cos x} \right)\].

Thus, substituting \[f\left( x \right)=1,g\left( x \right)=\cos x\] in the quotient rule of differentiation, we get \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{dy}{dx}\left( \dfrac{1}{\cos x} \right)=\dfrac{\cos x\times \dfrac{d}{dx}\left( 1 \right)-1\times \dfrac{d}{dx}\left( \cos x \right)}{{{\left( \cos x \right)}^{2}}}\]. \[...\left( 1 \right)\]

We know that differentiation of a constant is zero with respect to any variable. Thus, we have\[\dfrac{d}{dx}\left( 1 \right)=0\]. \[...\left( 2 \right)\]

We also know that differentiation of function of the form \[y=\cos x\] is \[\dfrac{dy}{dx}=-\sin x\]. \[...\left( 3 \right)\]

Substituting the value of equation \[\left( 2 \right), \left( 3 \right)\] in equation \[\left( 1 \right)\], we have \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{\cos x\times \dfrac{d}{dx}\left( 1 \right)-1\times \dfrac{d}{dx}\left( \cos x \right)}{{{\left( \cos x \right)}^{2}}}=\dfrac{\cos x\times 0-1\times \left( -\sin x \right)}{{{\cos }^{2}}x}=\dfrac{\sin x}{{{\cos }^{2}}x}\].

We know that \[\dfrac{\sin x}{\cos x}=\tan x\]. Thus, we have \[\dfrac{dy}{dx}\left( \sec x \right)=\dfrac{\sin x}{{{\cos }^{2}}x}=\dfrac{\tan x}{\cos x}=\tan x\sec x\].

Hence, the derivative of the function \[y=\sec x\] is \[\dfrac{dy}{dx}\left( \sec x \right)=\tan x\sec x\].

The derivative of any function \[y=f\left( x \right)\] with respect to variable \[x\] is a measure of the rate at which the value of the function changes with respect to the change in the value of variable \[x\]. The first derivative of any function also signifies the slope of the function when the graph of \[y=f\left( x \right)\] is plotted against \[x\] considering only real values of the function.

Note: It’s necessary to use quotient rules to find the derivative of the given function. We can also use the basic formula for finding the derivative of any function using limit.

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