# How do you differentiate #f(x)=tanx/(cosx-4)#?

Using the quotient rule, we get:

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To differentiate the function ( f(x) = \frac{\tan(x)}{\cos(x) - 4} ), you can use the quotient rule, which states that if ( u(x) ) and ( v(x) ) are differentiable functions, then the derivative of their quotient is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

Using this rule, the derivative of ( f(x) ) with respect to ( x ) is:

[ f'(x) = \frac{d}{dx} \left( \frac{\tan(x)}{\cos(x) - 4} \right) ]

[ = \frac{\sec^2(x)(\cos(x) - 4) - \tan(x)(-\sin(x))}{(\cos(x) - 4)^2} ]

[ = \frac{\sec^2(x)(\cos(x) - 4) + \tan(x)\sin(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\sec^2(x)\cos(x) - 4\sec^2(x) + \tan(x)\sin(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x)}{\cos^2(x) - 4} - \frac{4\sec^2(x)}{\cos^2(x) - 4} + \frac{\sin(x)\tan(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x)}{\cos^2(x) - 4} - \frac{4}{\cos^2(x) - 4} + \frac{\sin(x)\tan(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x)}{1 - 4\cos^2(x)} - \frac{4}{1 - 4\cos^2(x)} + \frac{\sin(x)\tan(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x)}{1 - 4\cos^2(x)} - \frac{4}{1 - 4\cos^2(x)} + \frac{\sin(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x) - 4}{1 - 4\cos^2(x)} + \frac{\sin(x)}{(\cos(x) - 4)^2} ]

[ = \frac{\cos(x) - 4 + \sin(x)(1 - 4\cos^2(x))}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) - 4 + \sin(x)(1 - 4\cos^2(x))}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) - 4 + \sin(x) - 4\sin(x)\cos^2(x)}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4 - 4\sin(x)\cos^2(x)}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x)\cos^2(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x)(1 - \sin^2(x)) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x) - 4\sin(x) + 4\sin^3(x) - 4}{(1 - 4\cos^2(x))^2} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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