CAIE A-Level Further Math 2.3.2 An expression for x y d

$\frac{\mathrm{d} y}{\mathrm{~d} t}=\mathrm{e}^{t}(t+1), \frac{\mathrm{d} x}{\mathrm{~d} t}=-t^{-2}$

B1

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{d} x}=-t^{2} \mathrm{e}^{t}(t+1)=-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)$

M1 A1

Uses chain rule, AG.

3

$\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)=-\mathrm{e}^{t}\left(3 t^{2}+2 t\right)-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)=-\mathrm{e}^{t}\left(t^{3}+4 t^{2}+2 t\right)$

M1 A1

Applies product rule.

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=t^{3} \mathrm{e}^{t}\left(t^{2}+4 t+2\right)$

M1 A1

Uses chain rule, multiplies by their
$\frac{\mathrm{d} t}{\mathrm{~d} x}$ for M 1 .

Alternative method for question 2(b)

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-\mathrm{e}^{t}\left(3 t^{2} \frac{\mathrm{~d} t}{\mathrm{~d} x}+2 t \frac{\mathrm{~d} t}{\mathrm{~d} x}\right)-\mathrm{e}^{t}\left(t^{3}+t^{2}\right) \frac{\mathrm{d} t}{\mathrm{~d} x}$

(M1 A1)

Differentiates $-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)$
implicitly.

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\mathrm{e}^{t}\left(-4 t^{2}-2 t-t^{3}\right) \frac{\mathrm{d} t}{\mathrm{~d} x}=t^{3} \mathrm{e}^{t}\left(t^{2}+4 t+2\right)$

(M1 A1)

Substitutes $\frac{\mathrm{d} t}{\mathrm{~d} x}=-t^{2}$.

4