How do you find the first and second derivative of #ln(x^4+5x^2)^(3/2) #?
By the chain and power rule we get
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By the power and chain rule we get
which simplifies to
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To find the first and second derivative of ( \ln{(x^4 + 5x^2)}^{\frac{3}{2}} ), you can follow these steps:
- First, let ( u = x^4 + 5x^2 ).
- Then, apply the chain rule to differentiate ( \ln{(u)}^{\frac{3}{2}} ) with respect to ( x ).
- After finding the first derivative, apply the chain rule again to find the second derivative.
The first derivative is:
[ \frac{d}{dx}\left[\ln{(x^4 + 5x^2)}^{\frac{3}{2}}\right] = \frac{3}{2} \cdot \frac{1}{(x^4 + 5x^2)} \cdot (x^4 + 5x^2)' ]
[ = \frac{3}{2} \cdot \frac{1}{(x^4 + 5x^2)} \cdot (4x^3 + 10x) ]
[ = \frac{3(4x^3 + 10x)}{2(x^4 + 5x^2)} ]
To find the second derivative, differentiate the first derivative with respect to ( x ).
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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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