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

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Answer 1 · 2018-06-02T20:28:16

#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)#

Answer 2 · 2018-06-02T20:37:47

#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#