Can someone pls help me find the second derivative of this. I need to find the concave up and down intervals but I’m having trouble getting the second derivative
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Starting from
[tex]f(x)=\dfrac14(x+1)^{\frac83}-4(x+1)^{\frac23}[/tex]
take the first derivative using the power and chain rules:
[tex]f'(x)=\dfrac83\cdot\dfrac14(x+1)^{\frac83-1}-\dfrac23\cdot4(x+1)^{\frac23-1}[/tex]
[tex]f'(x)=\dfrac23(x+1)^{\frac53}-\dfrac83(x+1)^{-\frac13}[/tex]
Now take the second derivative:
[tex]f''(x)=\dfrac53\cdot\dfrac23(x+1)^{\frac53-1}-\left(-\dfrac13\right)\cdot\dfrac83(x+1)^{-\frac13-1}[/tex]
[tex]f''(x)=\boxed{\dfrac{10}9(x+1)^{\frac23}+\dfrac89(x+1)^{-\frac43}}[/tex]
Optionally, you can condense the second derivative a bit by factoring out [tex]\frac89(x+1)^{-\frac43}[/tex], which gives
[tex]f''(x)=\dfrac89 (x+1)^{-\frac43} \left(80(x+1)^{\frac63}+1\right)[/tex]
[tex]f''(x)=\dfrac89 (x+1)^{-\frac43} \left(80(x+1)^2+1\right)[/tex]
[tex]f''(x)=\dfrac89 (x+1)^{-\frac43} \left(80x^2+160x+81\right)[/tex]