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P Pankaj Sanodiya
Given, Now, As we know the derivative of any function     Also chain rule of derivative, Hence the derivative of the given function is

P Pankaj Sanodiya
Given  Now, As we know  the Multiplication property of derivative, Also the property  Applying those properties we get,  the derivative of the given function is,

P Pankaj Sanodiya
Given  Now, As we know the derivative of any function

P Pankaj Sanodiya
Given, Now, As we know the derivative of any function   Now, As we know the derivative of any function   Hence the derivative of the given function is

P Pankaj Sanodiya
Given, Now, As we know the derivative of any function   Also the property  Applying those properties,we get

P Pankaj Sanodiya
Given, And the Multiplication property of derivative, Also the property  Applying those properties we get,

P Pankaj Sanodiya
Given, Now As we know the Multiplication property of derivative,(the product rule) And also the property Applying those properties we get,

P Pankaj Sanodiya
Given  Now, As we know the product rule  of derivative, The derivative of the given function is

P Pankaj Sanodiya
Given  Now, As we know, the Multiplication property of derivative, Hence derivative of the given function is:

P Pankaj Sanodiya

Given,

$f(x)=\frac{\sin ( x+a )}{ \cos x }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence the derivative of the given function is:

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\frac{d(\sin (x+a))}{dx})-\sin(x+a)(\frac{d(\cos x)}{dx})}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))-\sin(x+a)(-\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))+\sin(x+a)(\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (x+a-x)}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (a)}{(\cos x)^2}$

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P Pankaj Sanodiya

Given Function

$f(x)=\frac{a + b \sin x }{c+ d \cos x }$

Now, As we know the derivative of any function  of this type is:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence derivative of the given function will be:

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(\frac{d(a+b\sin x)}{dx})-(a+b\sin x)(\frac{d(c+d\cos x x)}{dx})}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(b\cos x)-(a+b\sin x)(d(-\sin x))}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+bd\cos^2 x+ad\sin x+bd\sin^2 x}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd(\sin^2 x+\cos^2 x)}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd}{(c+d\cos x)^2}$

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P Pankaj Sanodiya
Given, Now, As we know the chain rule of derivative, And, the property, Applying those properties, we get Hence Derivative of the given function is

P Pankaj Sanodiya
Given, which also can be written as  Now, As we know the derivative of such function   So, The derivative of the function is, Which can also be written as  .

P Pankaj Sanodiya
Given  Also can be written as which further can be written as Now,

P Pankaj Sanodiya
Given, Now, As we know the derivative of any function   Hence, The derivative of f(x) is  (Answer)

P Pankaj Sanodiya
Given,  the Multiplication property of derivative, Applying the property  Hence derivative of the function is .

P Pankaj Sanodiya
Given, Now, As we know the chain rule of derivative, Applying this property we get,

P Pankaj Sanodiya
Given Now, As we know the chain rule of derivative, And the Multiplication property of derivative, And, the property, Also the property  Applying those properties we get,

P Pankaj Sanodiya
Given Now, As we know the chain rule of derivative, And, the property, Also the property  applying those properties we get,

P Pankaj Sanodiya
Given It can also be written as  Now, As we know, the property, and the property  applying that property we get
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