How do you find #(d^2y)/(dx^2)# given #y=x^(9/4)#?

Question

Elijah Abram · Accepted Answer

The answer is #45/16*x^(1/4)# We use #(x^n)'=nx^(n-1)# #y=x^(9/4)# #dy/dx=9/4*x^(9/4-1)=9/4x^(5/4)# #(d^2x)/dy^2=9/4*5/4x^(5/4-1)=45/16*x^(1/4)#

Aurora Alvarado · Answer

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