Question 965782
1.The slope of the tangent line is the value of the derivative at the point of interest. 
{{{y=cos(x)}}}
{{{dy/dx=-sin(x)}}}
{{{m=dy/dx=-sin(pi/4)=-sqrt(2)/2}}}
Find the value of y at {{{x=pi/4}}}.
{{{y=cos(pi/4)}}}
{{{y=sqrt(2)/2}}}
Now use the point slope form of a line,
{{{y-sqrt(2)/2=-(sqrt(2))/2(x-pi/4)}}}
{{{y-sqrt(2)/2=-(sqrt(2)/2)x-(pi*sqrt(2))/8}}}
{{{highlight(y=-(sqrt(2)/2)x+(sqrt(2)/2)(1-pi/4))}}}
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*[illustration dae.JPG].
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2. {{{dy/dx=(5x+2)(3x^2+4)+(x^3+4x)(5)}}}
{{{dy/dx=(15x^3+6x^2+20x+8)+(5x^3+20x)}}}
{{{dy/dx=20x^3+6x^2+40x+8}}}
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3. {{{f(x)=1/(x-3)}}} for ({{{-infinity}}},{{{3}}})U({{{3}}},{{{infinity}}}) and
{{{f(x)=0}}} for {{{x=3}}}