How do you differentiate #V=4/3pi^3+8pir^2#?

Answer 1

Assuming #r# is the variable, it is #16πr#.

The power rule states that #(x^n)' = nx^(n-1)#. We also know that if #y# is a function of #x#, then for any real nonzero constant #c#, #(cy)' = c(y)'#. Also recall that the derivative of a constant by itself is zero, and that the derivative of a sum is the sum of derivatives of the terms.
Since #4/3π^3# is a constant, its derivative is #0#.
#(8πr^2)' = 8π(r^2)' = 8π(2r) = 16πr#.

Therefore,

#(dV)/(dr) = 0 + 16πr = 16πr#.
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Answer 2

To differentiate ( V = \frac{4}{3}\pi r^3 + 8\pi r^2 ) with respect to ( r ), use the power rule for differentiation:

[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3}\pi r^3 + 8\pi r^2 \right) = \frac{d}{dr} \left( \frac{4}{3}\pi r^3 \right) + \frac{d}{dr} \left( 8\pi r^2 \right) ]

Now, differentiate each term separately:

[ \frac{d}{dr} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi \frac{d}{dr} (r^3) = \frac{4}{3}\pi \times 3r^2 = 4\pi r^2 ]

[ \frac{d}{dr} \left( 8\pi r^2 \right) = 8\pi \frac{d}{dr} (r^2) = 8\pi \times 2r = 16\pi r ]

So, the derivative of ( V ) with respect to ( r ) is:

[ \frac{dV}{dr} = 4\pi r^2 + 16\pi r ]

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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.

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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