How do you find the derivatives of #z=ln((y^4+1)^2/(y-1)^3)#?

Answer 1

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Answer 2

To find the derivative of ( z = \ln\left(\frac{{(y^4 + 1)^2}}{{(y - 1)^3}}\right) ) with respect to ( y ), you would use the chain rule and the quotient rule. The steps are as follows:

  1. Apply the quotient rule: [ \frac{{d}}{{dy}}\left(\ln\left(\frac{{(y^4 + 1)^2}}{{(y - 1)^3}}\right)\right) = \frac{{\frac{{d}}{{dy}}\left((y^4 + 1)^2\right) \cdot (y - 1)^3 - (y^4 + 1)^2 \cdot \frac{{d}}{{dy}}\left((y - 1)^3\right)}}{{((y - 1)^3)^2}} ]

  2. Differentiate the terms using the chain rule and the power rule: [ \frac{{d}}{{dy}}\left((y^4 + 1)^2\right) = 2(y^4 + 1)^2 \cdot \frac{{d}}{{dy}}(y^4 + 1) = 2(y^4 + 1)^2 \cdot 4y^3 ] [ \frac{{d}}{{dy}}\left((y - 1)^3\right) = 3(y - 1)^2 \cdot \frac{{d}}{{dy}}(y - 1) = 3(y - 1)^2 \cdot 1 ]

  3. Substitute the derivatives back into the quotient rule expression.

  4. Simplify the expression if possible.

So, the derivative of ( z ) with respect to ( y ) is: [ \frac{{8y^3(y^4 + 1)^2 - 3(y - 1)^2(y^4 + 1)^2}}{{(y - 1)^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.

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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