Derivative of x= (2t + 3)^(3/2)/3 Doesn't Make Sense to Me

[BigNate]

Hello Everyone,

I can't figure out the derivative to the following:

x= (2t + 3)^(3/2)/3

Correct answer: dx/dt = (2t + 3)^(1/2)
My INCORRECT answer: 1/2 (2t + 3)^(1/2)

For some reason, the correct solution has another 2 that will cancel with the 1/2 I get at the end. I cannot figure out where this 2 is coming from.

Thank you in advance for your help and for your time.

 

[Deleted member 4993]

Hello Everyone,

I can't figure out the derivative to the following:

x= (2t + 3)^(3/2)/3

Correct answer: dx/dt = (2t + 3)^(1/2)
My INCORRECT answer: 1/2 (2t + 3)^(1/2)

For some reason, the correct solution has another 2 that will cancel with the 1/2 I get at the end. I cannot figure out where this 2 is coming from.

Thank you in advance for your help and for your time.

Repeated (and careful) application of chain-rule

Let

f₁(t) = (2t + 3)

d[f₁(t)]/dt = 2

f₂(t) = [f₁(t)]^(3/2)

d[f₂(t)]/dt = 3/2 * [f₁(t)]^(1/2) * d[f₁(t)]/dt = 3/2 * [2t + 3]^(1/2) * 2 = 3 * [2t + 3]^(1/2)

d[f₂(t)/3]/dt = 1/3 * d[f₂(t)]/dt = 1/3 * [3 * [2t + 3]^(1/2)] = [2t + 3]^(1/2)

 

[stapel]

I can't figure out the derivative to the following:

x= (2t + 3)^(3/2)/3

Correct answer: dx/dt = (2t + 3)^(1/2)
My INCORRECT answer: 1/2 (2t + 3)^(1/2)

Unfortunately, it is not possible for us to check work which we cannot see. In future, kindly please provide this information. Thank you!

 

[BigNate]

Thank you both for your help. It was the chain rule that had me confused. Once I understood proper application of the chain rule, I was fine.