How do you solve the differential #dy/dx=(10x^2)/sqrt(1+x^3)#?
We know that
so, grouping variables
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To solve the differential equation ( \frac{dy}{dx} = \frac{10x^2}{\sqrt{1+x^3}} ), you can separate variables and integrate both sides. Here's the stepbystep process:

Separate variables: ( \frac{dy}{\sqrt{1+y^3}} = 10x^2 , dx )

Integrate both sides: ( \int \frac{1}{\sqrt{1+y^3}} , dy = \int 10x^2 , dx )

For the left integral, you can use the substitution ( u = 1+y^3 ), so ( du = 3y^2 , dy ): ( \frac{1}{3} \int \frac{1}{\sqrt{u}} , du = \int 10x^2 , dx )

After integrating both sides: ( \frac{2}{3} \sqrt{u} = \frac{10}{3} x^3 + C )

Substitute ( u = 1+y^3 ) back in: ( \frac{2}{3} \sqrt{1+y^3} = \frac{10}{3} x^3 + C )

Solve for ( y ): ( \sqrt{1+y^3} = 5x^3 + C' )

Square both sides to isolate ( y ): ( 1+y^3 = (5x^3 + C')^2 )

Subtract 1 from both sides: ( y^3 = (5x^3 + C')^2  1 )

Finally, solve for ( y ) by taking the cube root: ( y = \sqrt[3]{(5x^3 + C')^2  1} )
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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