How do you find #(d^2y)/(dx^2)# for #5x^3=-4y^2+4#?
Differentiating:
Differentiate again. This time the quotient rule will be used!
Simplify:
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To find (d^2y)/(dx^2) for the equation 5x^3 = -4y^2 + 4, you first need to find the first derivative dy/dx by implicit differentiation, then differentiate dy/dx with respect to x again to find (d^2y)/(dx^2).
Step 1: Implicit differentiation of the equation 5x^3 = -4y^2 + 4 with respect to x gives: 15x^2 = -8yy'
Step 2: Solve for y': y' = (15x^2) / (-8y)
Step 3: Differentiate y' obtained in step 2 with respect to x to find (d^2y)/(dx^2): (d^2y)/(dx^2) = d/dx [(15x^2) / (-8y)] (d^2y)/(dx^2) = [(30x)(-8y) - (15x^2)(-8(dy/dx))] / (-8y)^2 (d^2y)/(dx^2) = (240xy + 120x^2(dy/dx)) / (64y^2)
Now, substitute the value of dy/dx from step 2 into this expression: (d^2y)/(dx^2) = (240xy + 120x^2((15x^2) / (-8y))) / (64y^2) (d^2y)/(dx^2) = (240xy - 225x^4) / (64y^3)
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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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