Differentiate the function. y = 3e^x + 4/3sqrt x

Question

Please help

Answer 1

1 / √x      is the same as      x^-1/2 .

 

 

y = 3e^x + (4 / 3)x^-1/2

 

 

We use a combination of power rule and product rule.

 

dy/dx = 3e^x - (1 / 2)(4 / 3)x^{(-1/2 - 1)}

 

dy/dx = 3e^x - (2 / 3)x^-3/2

 

dy/dx = 3e^x - (2 / 3)(1 / √(x³))

 

dy/dx = 3e^x - (2 / 3)(1 / (x√x))

 

dy/dx = 3e^x - (2 / (3x√x))

 

Answer 2

I believe it is this since when you differentiate e^x the answer is e^x and differentiate sqrtx it's 1/2 * 1/sqrt x

 

so it should be 

 

dy/dx= 3e^x + 2/3 * 1/sqrt x

 

But I'd double check with someone who has taken calculus in the past decade since I haven't in over 25 years:) But I'm pretty sure that's right.