document.write( "Question 1193312: Determine the derivative of the following: \r
\n" ); document.write( "\n" ); document.write( "1.y= ln^3(x +sin x)\r
\n" ); document.write( "\n" ); document.write( "2.y= tan^-1(3x) \n" ); document.write( "
Algebra.Com's Answer #848655 by ElectricPavlov(122)\"\" \"About 
You can put this solution on YOUR website!
**1. y = ln^3(x + sin(x))**\r
\n" ); document.write( "\n" ); document.write( "* **Chain Rule:**
\n" ); document.write( " * Let u = ln(x + sin(x))
\n" ); document.write( " * Then y = u^3
\n" ); document.write( " * dy/du = 3u^2
\n" ); document.write( " * du/dx = 1 / (x + sin(x)) * (1 + cos(x)) \r
\n" ); document.write( "\n" ); document.write( "* **Apply Chain Rule:**
\n" ); document.write( " * dy/dx = dy/du * du/dx
\n" ); document.write( " * dy/dx = 3u^2 * (1 + cos(x)) / (x + sin(x)) \r
\n" ); document.write( "\n" ); document.write( "* **Substitute back u:**
\n" ); document.write( " * dy/dx = 3[ln(x + sin(x))]^2 * (1 + cos(x)) / (x + sin(x))\r
\n" ); document.write( "\n" ); document.write( "**2. y = tan^(-1)(3x)**\r
\n" ); document.write( "\n" ); document.write( "* **Chain Rule:**
\n" ); document.write( " * Let u = 3x
\n" ); document.write( " * Then y = tan^(-1)(u)
\n" ); document.write( " * dy/du = 1 / (1 + u^2)
\n" ); document.write( " * du/dx = 3\r
\n" ); document.write( "\n" ); document.write( "* **Apply Chain Rule:**
\n" ); document.write( " * dy/dx = dy/du * du/dx
\n" ); document.write( " * dy/dx = 1 / (1 + u^2) * 3 \r
\n" ); document.write( "\n" ); document.write( "* **Substitute back u:**
\n" ); document.write( " * dy/dx = 3 / (1 + (3x)^2)
\n" ); document.write( " * dy/dx = 3 / (1 + 9x^2) \r
\n" ); document.write( "\n" ); document.write( "I hope this helps! Let me know if you have any other questions.
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