Multiply the polynomials: (y – 8) and (3y – 4)

Explanation

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a. `(dy)/(dx)=(dy//dt)/(dx//dt)=(12t^(2)-6t-18)/(5t^(4)-15t^(2)-20)`
`"or "(dy)/(dx):|_(t=1)=(12-6-18)/(5-15-20)=(2)/(5)`
`"or "-5(dy)/(dx):|_(t=1)=-2" at "t=1`
b. Let us take
`P(x)=a(x-2)^(4)+b(x-2)^(3)+c(x-2)^(2)+d(x-2)-1`
`therefore" "P(2)=-1`
`0=P'(2)=d`
`2=P''(2)=2 c or c=1`
`-12=P'''(2)=6 b or b=-2`
`24=P^(iv)(2)=24a or a=1`
Thus, `P''(x)=12 (x-2)^(2)-12(x-2)+2`
`"or "P''(3)=12-12(1)+2=2`
c. `"Here, "sqrt((1+y^(4)))=sqrt((1+(1)/(x^(4))))=(sqrt(1+x^(4)))/(x^(2))" "(becausey=(1)/(x))`
`"or "(sqrt(1+y^(4)))/(sqrt(1+x^(4)))=(1)/(x^(2))" (1)"`
`"But "y=(1)/(x)`
`therefore" "(dy)/(dx)=-(1)/(x^(2))" (2)"`
`"From (1) and (2), "(sqrt(1+y^(4)))/(sqrt(1+x^(4)))=-(dy)/(dx)`
`"or "((dy)/sqrt(1+y^(4)))/((dx)/sqrt(1+x^(4)))=-1`
d. Obviously, f(x) is a linear function.
Also, from f'(0) = p and f(0) = q, f(x) = px +q
`"or "f''(0)=0`