1. Graph y = tan(x) and its derivative together on (-%u03C0/2, %u03C0/2). Does t

Question

1. Graph y = tan(x) and its derivative together on (-%u03C0/2, %u03C0/2).  Does the graph of the tangent function appear to have a smallest slope? A largest slope?  Is the slope ever negative? Give reasons for you answers?

2. Plot f(x) =ln(x) and f%u2019(x) = 1/x in your      calculator.  Remembering that the      derivative of f(x) is the instantaneous rate of change of f(x) why does it      make sense that 1/x is the derivative of ln(x)?

3. Find the derivative of;

f(x)= (x^3e^x)/2x-9                                                                         

Explanation / Answer

Possible derivation: d/dx((x^3 e^x)/(2 x-9)) Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = e^x x^3 and v = 2 x-9: = ((-9+2 x) (d/dx(e^x x^3))-e^x x^3 (d/dx(-9+2 x)))/(-9+2 x)^2 Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = e^x and v = x^3: = (-(e^x x^3 (d/dx(-9+2 x)))+(-9+2 x) x^3 d/dx(e^x)+e^x d/dx(x^3))/(-9+2 x)^2 The derivative of e^x is e^x: = (-(e^x x^3 (d/dx(-9+2 x)))+(-9+2 x) (e^x (d/dx(x^3))+e^x x^3))/(-9+2 x)^2 Use the power rule, d/dx(x^n) = n x^(n-1), where n = 3: d/dx(x^3) = 3 x^2: = (-(e^x x^3 (d/dx(-9+2 x)))+(-9+2 x) (e^x x^3+3 x^2 e^x))/(-9+2 x)^2 Differentiate the sum term by term and factor out constants: = ((-9+2 x) (3 e^x x^2+e^x x^3)-d/dx(-9)+2 d/dx(x) e^x x^3)/(-9+2 x)^2 The derivative of -9 is zero: = ((-9+2 x) (3 e^x x^2+e^x x^3)-e^x x^3 (2 (d/dx(x))+0))/(-9+2 x)^2 Simplify the expression: = ((-9+2 x) (3 e^x x^2+e^x x^3)-2 e^x x^3 (d/dx(x)))/(-9+2 x)^2 The derivative of x is 1: Answer: | | = ((-9+2 x) (3 e^x x^2+e^x x^3)-1 2 e^x x^3)/(-9+2 x)^2

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