SOLUTION: Determine the derivative of the following: 1.y= ln^3(x +sin x) 2.y= tan^-1(3x)

Question 1193312: Determine the derivative of the following:
1.y= ln^3(x +sin x)
2.y= tan^-1(3x)
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**1. y = ln^3(x + sin(x))**
* **Chain Rule:**
* Let u = ln(x + sin(x))
* Then y = u^3
* dy/du = 3u^2
* du/dx = 1 / (x + sin(x)) * (1 + cos(x))
* **Apply Chain Rule:**
* dy/dx = dy/du * du/dx
* dy/dx = 3u^2 * (1 + cos(x)) / (x + sin(x))
* **Substitute back u:**
* dy/dx = 3[ln(x + sin(x))]^2 * (1 + cos(x)) / (x + sin(x))
**2. y = tan^(-1)(3x)**
* **Chain Rule:**
* Let u = 3x
* Then y = tan^(-1)(u)
* dy/du = 1 / (1 + u^2)
* du/dx = 3
* **Apply Chain Rule:**
* dy/dx = dy/du * du/dx
* dy/dx = 1 / (1 + u^2) * 3
* **Substitute back u:**
* dy/dx = 3 / (1 + (3x)^2)
* dy/dx = 3 / (1 + 9x^2)
I hope this helps! Let me know if you have any other questions.

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