Question

If $y=a{t}^2+2bt+c$ and $t=a{x}^2+2bx+c$, then $\frac{d^{3}y}{d{x}^3}$ equals

• A. $24a^2(at+b)$
• B. $24a(ax+b{)}^2$
• C. $24a(at+b{)}^2$
• D. $24a^2(ax+b)$

Answer 1

The correct option is D $24a^2(ax+b)$

$y=a{t}^2+2bt+candt=a{x}^2+2bx+c$

$\frac{dy}{dx}=2at{t}^{\prime }+2b{t}^{\prime }$

$\frac{dt}{dx}=t^{\prime }=2ax+2b=2(ax+b)$

$\frac{dy}{dx}=2a(a{x}^2+2bx+c)2(ax+b)+2b.2(ax+b)$

$=4(ax+b)(a^2{x}^2+2abx+ac+b)$

$=4(a^3{x}^3+2a^{2}b{x}^2+(ac+b).ax+a^2{x}^{2}b+2a{b}^{2}x+abc+b^2)$

$\frac{d^{2}y}{d{x}^2}=4(3a^3{x}^2+2a^{2}b(2x\left)\right)+(ac+b).a+a^2\left(2x\right)b+2a{b}^2)$

$\frac{d^{3}y}{d{x}^3}=4(6a^{3}x+6a^{2}b)=24a^2(ax+b)$