How do you differentiate #y= -cos^-1 (1/x^5)#?
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To differentiate ( y = -\cos^{-1}\left(\frac{1}{x^5}\right) ), we'll use the chain rule. The derivative of (\cos^{-1}(u)) with respect to (u) is (-\frac{1}{\sqrt{1-u^2}}), and the derivative of (u = \frac{1}{x^5}) with respect to (x) is (-\frac{5}{x^6}). Applying the chain rule, we get:
[ \frac{dy}{dx} = -\frac{d}{dx}\left(\cos^{-1}\left(\frac{1}{x^5}\right)\right) = -\left(-\frac{1}{\sqrt{1-\left(\frac{1}{x^5}\right)^2}}\right) \times \left(-\frac{5}{x^6}\right) ]
[ = \frac{5}{x^6\sqrt{1-\frac{1}{x^{10}}}} ]
So, the derivative of ( y = -\cos^{-1}\left(\frac{1}{x^5}\right) ) is ( \frac{5}{x^6\sqrt{1-\frac{1}{x^{10}}}} ).
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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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