Question 6.35: Differentiate each of the following: (a) y = x²e^x (b) P = 2......
Differentiate each of the following:
(a) y = x²e^x (b) P = 2 \sqrt {Q}(Q + 5) (c) C = (Y + 4) ln(Y )
(a) Stage 1
State u and v and find du/dx and dv/dx.
y = x²e^x
u = x² and v = e^x
\frac {du}{dx} = 2x and \frac {dv}{dx} = e^x
Stage 2
Fill in the product rule, equation (6.32)
{\frac{\mathrm{d}y}{\mathrm{d}x}}=v{\frac{\mathrm{d}u}{\mathrm{d}x}}+u{\frac{\mathrm{d}v}{\mathrm{d}x}}= (e^x )(2x) + (x²)(e^x )
= e^x (2x + x²)
= xe^x (2 + x)
Result
y = x²e^x , then dy/dx = xe^x (2 + x)
(b) Stage 1
State u and v and find du/dQ and dv/dQ.
P = 2 \sqrt {Q}(Q + 5)
= 2(Q)^{1/2}(Q + 5)
u = 2Q^{(0.5)} and v = Q + 5
{\frac{\mathrm{d}u}{\mathrm{d}Q}}=2(0.5Q^{0.5-1}) and {\frac{\mathrm{d}v}{{d}{Q}}}=1
= 2(0.5Q^{−0.5})
= Q^{−0.5}
Stage 2
Fill in the product rule, equation (6.32)
{\frac{\mathrm{d}P}{\mathrm{d}Q}}= v{\frac{\mathrm{d}u}{\mathrm{d}Q}}+u{\frac{\mathrm{d}v}{\mathrm{d}Q}}= (Q + 5)(Q^{−0.5}) + (2 Q^{0.5})1
= Q^{0.5} + 5Q^{−0.5} + 2 Q^{0.5}
= 3Q^{0.5} + 5Q^{−0.5}
Result
P = 2\sqrt {Q}(Q + 5) then
\frac{dP}{dQ}= 3Q^{0.5} + \frac {5}{Q^{0.5}}(c) Stage 1
State u and v and find du/dY and dv/dY.
C = (Y + 4) ln(Y )
u = (Y + 4) and v = ln(Y )
\frac {du}{dY} = 1 and \frac {dv}{dY} = \frac {1}{Y}
Stage 2
Fill in the product rule, equation (6.32)
{\frac{\mathrm{d}C}{\mathrm{d}Y}}= v{\frac{\mathrm{d}u}{\mathrm{d}Y}}+u{\frac{\mathrm{d}v}{\mathrm{d}Y}}= ln(Y )(1) + (Y + 4) \frac {1}{Y}
= ln(Y ) + \frac {Y + 4}{Y}
= ln(Y ) + 1 + \frac {4}{Y}
Result
C = (Y + 4) ln(Y ) \frac {dC}{dY} = ln(Y ) + 1 + \frac {4}{Y}