How do you use the chain rule to differentiate #y=(3x^4-7x^3+3x^2-5x)^3#?

Answer 1

#(dy)/(dx)=3(12x^3-21x^2+6x-5)(3x^4-7x^3+3x^2-5x)^2#

Chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#
Let #u=3x^4-7x^3+3x^2-5x#, then #(du)/(dx)=12x^3-21x^2+6x-5#
#y=u^3#, so #(dy)/(du)=3u^2=3(3x^4-7x^3+3x^2-5x)^2#
#:.(dy)/(dx)=(12x^3-21x^2+6x-5)*3(3x^4-7x^3+3x^2-5x)^2=3(12x^3-21x^2+6x-5)(3x^4-7x^3+3x^2-5x)^2#
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Answer 2

To differentiate ( y = (3x^4 - 7x^3 + 3x^2 - 5x)^3 ) using the chain rule, you first find the derivative of the outer function, which is raising to the power of 3. Then you multiply by the derivative of the inner function, which is (3x^4 - 7x^3 + 3x^2 - 5x), using the power rule and the sum/difference rule for derivatives. So the derivative is:

[ \frac{dy}{dx} = 3(3x^4 - 7x^3 + 3x^2 - 5x)^2 \times (12x^3 - 21x^2 + 6x - 5) ]

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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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