##### I don't even know where to begin with this second-degree derivative problem.

 Calculus Tutor: None Selected Time limit: 1 Day

Aug 10th, 2015

Start by taking the first derivative of the function.  To do this you must use the chain rule.  Remember that the chain rule says we take the derivative of the outside function (sin), leave the inside (2x + 3) the same, then multiply by the derivative of the inside.

y = sin(2x + 3)

dy/dx = cos(2x + 3) * 2

dy/dx = 2 cos(2x + 3)

Now, we can find the second derivative by applying the chain rule again.

d^2y / dx^2 = 2 * -sin (2x + 3) *2

d^2y / dx^2 = - 4 sin(2x + 3)

Since we have determined the second derivative, lets take a look at the statement we are trying to prove.

d^2y / dx^2 + 4y = 0

Now put the second derivative and the original function into the correct places in the equation

-4 sin(2x + 3) + 4 * sin(2x +3) = 0

Now since the sine functions are the same (have the same things in parenthesis), we can add the two terms together.

(-4 + 4) sin(2x + 3) = 0

0 = 0

I hope that this makes sense to you.

Please let me know if you need any clarification or anything explained in more detail. I'm always happy to answer your questions.
Aug 10th, 2015

Oh, there we go, I was missing the substitution of the original equation.

Aug 10th, 2015

When you are asked to prove something you always need to look back to the original equation.  So make sure that you have all the parts to make a substitution and from there the solution should be pretty apparent.

Aug 10th, 2015

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Aug 10th, 2015
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Aug 10th, 2015
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