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How do you differentiate #f(x)= 5/x^3 - 3/ x^(1/3)#?

Answer 1

Emma Aldridge

#f^'(x) = -15/x^4 + 1/x^(4/3)#

Put it on root notation

#f(x) = 5x^(-3) - 3x^(-1/3)#

Using the polynomial formula, #f^'(x) = nx^(n-1)#, we have

#f^'(x) = 5*(-3)* x^(-3-1) -3*(-1/3)*x^(-1/3-1)#

#f^'(x) = -15x^(-4) + x^(-4/3)#

Since the problem started with fraction notation, let us put the answer back in that notation

#f^'(x) = -15/x^4 + 1/x^(4/3)#

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Answer 2

Paisley Johnson

To differentiate the function ( f(x) = \frac{5}{x^3} - \frac{3}{x^{1/3}} ), you can use the power rule and the constant multiple rule for differentiation.

[ f'(x) = \frac{d}{dx}\left(\frac{5}{x^3}\right) - \frac{d}{dx}\left(\frac{3}{x^{1/3}}\right) ]

[ = 5 \cdot \frac{d}{dx}\left(\frac{1}{x^3}\right) - 3 \cdot \frac{d}{dx}\left(\frac{1}{x^{1/3}}\right) ]

Using the power rule ( \frac{d}{dx}\left(\frac{1}{x^n}\right) = -\frac{n}{x^{n+1}} ):

[ = 5 \cdot \left(-3 \cdot \frac{1}{x^{3+1}}\right) - 3 \cdot \left(-\frac{1}{3x^{1/3+1}}\right) ]

[ = -\frac{15}{x^4} + \frac{1}{x^{4/3}} ]

Thus, the derivative of ( f(x) ) is ( f'(x) = -\frac{15}{x^4} + \frac{1}{x^{4/3}} ).

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Answer from HIX Tutor

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.