How do you find the first and second derivative of #ln(x^4+5x^2)^(3/2) #?

Answer 1

#f'(x)=3/2*(10*x+4x^3)/(5x^2+x^4)#
#f''(x)=-(3(25+5x^2+2x^4))/(x^2*(5+x^2)^2)#

By the chain and power rule we get

#f'(x)=1/(x^4+5x^2)^(3/2)*3/2*(x^4+5x^2)^(1/2)*(4x^3+10x)# which can be simplified to
#f'(x)=(3(10x+4x^3))/(2*(5x^2+x^4))# By the quotient rule we get
#f''(x)=3/2*((10+12x^2)(5x^2+x^4)-(10x+4x^3)*(10x+4x^3))/(5x^2+x^4)^2)# which can be simplified to
#f''(x)=-(3(25+5x^2+2x^4))/(x^2*(5+x^2)^2)#
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Answer 2

#f'(x)=(3*(10x+4x^3))/(2*(5x^2+x^4))#
#f''(x)=-(3*25+5x^2+2x^4)/(x^2*(5+x^2)^2)#

By the power and chain rule we get

#f'(x)=1/(x^4+5x^2)^(3/2)*(3/2)(x^4+5x^2)^(1/2)*(4x^3+10x)#

which simplifies to

#f'(x)=(3(10x+4x^3))/(2(5x^2+x^4))# By the quotient rule we get #f''(x)=3/2*((10+12x^2)*(5x^2+x^4)-(10x+4x^3)*(10x+4x^3))/(5x^2+x^4)^2#
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Answer 3

To find the first and second derivative of ( \ln{(x^4 + 5x^2)}^{\frac{3}{2}} ), you can follow these steps:

  1. First, let ( u = x^4 + 5x^2 ).
  2. Then, apply the chain rule to differentiate ( \ln{(u)}^{\frac{3}{2}} ) with respect to ( x ).
  3. 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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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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