^{x}cosx

^{3}

^{2})

^{-5}

^{2})

^{2}(x

^{2})

^{x})

^{lnx}

1. Find the derivative f'(x).

(a) f(x)=xe^{x}cosx

(b) f(x)=secxtanx

(c) f(x)=sinx/1+cosx

2. Let g(x) be a differentiable function such that g(0)=2 and g'(0)=3.

(a) f(x)=(g(x))^{3}

(b) f(x)=g(7x)

3. If h(t)=300+12sin(t/2), find h"(pi/3).

4. Find the derivative f'(x).

(a) f(x)=(4-2x-3x^{2})^{-5}

(b) f(x)=x(square root of 1+x^{2})

(c) f(x)=cos^{2}(x^{2})

5. Find the derivative f'(x).

(a) f(x)=lnx/x

(b) f(x)=arctan(e^{x})

(c) f(x)=x^{lnx}

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Marked as Best Answer

I'll do part (a) of each problem and give somebody else the chance to do the rest. :)

1a. Use the product rule, (uv)'=u'v+uv'.

For a product of 3 functions, (uvw)'=u'vw+uv'w+uvw'

(xe^{x}cosx)'=e^{x}cosx+xe^{x}cosx-xe^{x}sinx

2a. Use the chain rule, f(g(x))'=f'(g(x)) g'(x).

(g(x)³)'=3g(x)² g'(x)

(g³)'(0)=3g(0)² g'(0)=3*2²*3=36

3. h(t)=300+12sin(t/2)

h'(t)=6cos(t/2)

h''(t)=-3sin(t/2)

h''(Pi/3)=-3sin(Pi/6)=-3/2

4a. Chain rule again: outside derivative * inside derivative

f(x)=(4-2x-3x²)^{-5}

f'(x)=-5(4-2x-3x²)^{-6}*(-2-6x)

5a. Quotient rule or product rule, I'll use the product rule:

(ln(x)/x)'=(x^{-1}*ln(x))' =-x^{-2}ln(x)+x^{-1}*x^{-1}=-x^{-2}ln(x)+x^{-2}

Good luck on your exam! :)

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