# Homework # 4 (The Chain Rule)
# OPMT 5701
# Fall 2006
# 
# Problem 1:
> y:=1/(2-3*x)^4;
> y_prime:=diff(y,x);
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# 
# Problem 2:
> y:=1/(4*x^2-3)^(1/3);
> y_prime:=diff(y,x);
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# Problem 3:
> y:=sqrt(sqrt(x+1));
> y_prime:=diff(y,x);
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# Problem 4:
> y:=x^2*(2*x-5)^5;
> y_prime:=diff(y,x);
> factor(%);
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# 
# Problem 5:
> y:=(x^2+9)*sqrt(x^2+4);
> y_prime:=diff(y,x);
> simplify(%);
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# Problem 6:
> y:=((x+2)/(x-3))^4;
> y_prime:=diff(y,x);
> simplify(%);
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# Problem 7:
> f:=(x+1)^2*(x+2)^3;
> f_prime:=diff(f,x);
> factor(%);
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# 
# Multiple Choice:
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# Problem 8:
> y:=sqrt(x^2-9);
> yprime:=diff(y,x);
> slope:=subs(x=5,yprime);
> simplify(slope);
# So, we know that the slope of the tangent line is: 5/4.  Let us now plug in a point on the line:
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> y0:=subs(x=5,y);
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> y_tangent:=slope*(x-5)+y0;
> simplify(%);
# as we can tell from the above equation for the line, the answer is E) y = 5*(`*`(x))/(`*`(4))+`-`(`*`(9)/(`*`(4)));.
# 
# Problem 9:
> g:=x^4*(2*x-1)^10;
> g_prime:=diff(g,x);
> slope:=subs(x=1,g_prime);
# so, the answer is B) 24.
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# Problem 10:
> y:=u^5-8*u^2+2*u-1;
> u:=sqrt(x+10);
> y_prime:=diff(y,x);
> slope:=subs(x=-9,y_prime);
# so, the answer is B) `+`(`-`(`*`(9)/(`*`(2))));.
> 
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