How do you differentiate #f(x)= tanx# twice using the quotient rule?
Thus, in regards to your role:
Next, in line with the chain rule:
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To differentiate ( f(x) = \tan(x) ) twice using the quotient rule, first, differentiate ( f(x) ) once to find ( f'(x) ), then differentiate ( f'(x) ) to find ( f''(x) ).
The function ( f(x) = \tan(x) ) can be expressed as the quotient of two functions: ( f(x) = \frac{\sin(x)}{\cos(x)} ).
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First differentiation: Apply the quotient rule: [ f'(x) = \frac{\cos(x) \cdot \cos(x) - (-\sin(x)) \cdot \sin(x)}{\cos^2(x)} ] [ f'(x) = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} ] Since ( \cos^2(x) + \sin^2(x) = 1 ): [ f'(x) = \frac{1}{\cos^2(x)} ]
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Second differentiation: Apply the quotient rule to ( f'(x) = \frac{1}{\cos^2(x)} ): [ f''(x) = \frac{-2\sin(x)(-\sin(x))}{\cos^3(x)} ] [ f''(x) = \frac{2\sin^2(x)}{\cos^3(x)} ] Since ( \sin^2(x) = 1 - \cos^2(x) ): [ f''(x) = \frac{2(1 - \cos^2(x))}{\cos^3(x)} ] [ f''(x) = \frac{2 - 2\cos^2(x)}{\cos^3(x)} ]
You can simplify this expression further, but this is the second derivative of ( f(x) = \tan(x) ) obtained using the quotient rule twice.
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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.
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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