Question 200580
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From algebra, you need to know that 
1. {{{sqrt(u)=u^(1/2)}}} 
2. {{{1/(u^n) = u^(-n)}}}
and from calculus you need to know that

3. For {{{y=u^n}}}, then {{{(dy)/(dx)=n*u^(n-1)((du)/(dx))}}}

4. For {{{y=cos(u)}}}, then {{{ (dy)/(dx) = -sin(u)((du)/(dx))}}}

5. For {{{y=e^u}}}, then {{{(dy)/(dx)=(e^u)((du)/(dx))}}}

6. For {{{y=ln(u)}}}, then {{{(dy)/(dx)=((du)/(dx))/u}}}

7. For {{{y=u*v}}}, then {{{(dy)/(dx)=u((dv)/(dx))+v((du)/(dx))}}}


1. {{{y=3x^2 + 4x + 2}}}
   {{{(dy)/(dx)=6x+4}}} 

2. {{{y=x^2*sqrt(x) + 1/sqrt(x) - cos(3x)}}}
   {{{y=(x^2)(x^(1/2)) + 1/(x^(1/2)) - cos(3x)}}}
   {{{y=x^(2+1/2)+x^(-1/2)-cos(3x)}}}
   {{{y=x^(5/2)+x^(-1/2)-cos(3x)}}}
   {{{(dy)/(dx)=(5/2)x^(5/2-1)+(-1/2)x^(-1/2-1)-(-sin(3x)*3)}}}
   {{{(dy)/(dx)=(5/2)x^(3/2)-(1/2)x^(-3/2)+3sin(3x)}}}

3. {{{y=e^(3x) (x^2 + ln(x))}}}
   {{{(dy)/(dx)=e^(3x)(2x+1/x)+(x^2+ln(x))(e^(3x))(3)}}} 
   {{{(dy)/(dx)=e^(3x)(2x+1/x)+3e^(3x)(x^2+ln(x))}}}
   {{{(dy)/(dx)=e^(3x)((2x+1/x)+3(x^2+ln(x)))}}}
   {{{(dy)/(dx)=e^(3x)(2x+1/x+3x^2+3*ln(x))}}}


Edwin</pre>