Question

b) V 0-20303 dVIdQ dT dR c) D-10 6s2 +3s4/5 g) B=30' (20-30°) dB de

How do I find the derivatives for these functions without using the product rule?

Answer 1

A. $\large J = 2K^2 + 4K^3 - 3$

Differentiating J wrt K we get

$\large \frac{dJ}{dK} = \frac{d(2K^2 + 4K^3 - 3)}{dK}$

$\large \frac{dJ}{dK} =2 \frac{dK^2}{dK} + 4\frac{dK^3}{dK} - \frac{d 3}{dK}$

$\large \frac{dJ}{dK} =4K + 12K^2$

b. $\large V = Q^2 - 2Q + 3Q^3$

Differentiating V wrt Q we get

$\large \frac{dV}{dQ} = \frac{d(Q^2 - 2Q + 3Q^3)}{dQ}$

Similarly as we have done in previous case

$\large \frac{dV}{dQ} = 2Q - 2 + 9Q^2$

C. $\large D = 10 - 6S^2 + 3S^{\frac{4}{3}}$

Differentiating D wrt S we get

ds dS

$\large \frac{dD}{dS}= - 12S + 4S^{\frac{1}{3}}$

D. $\large Y = 16 - 4X^{\frac{1}{2}} + 6X^{\frac{-1}{2}}$

$\large \frac{dY}{dX} = \frac{d(16 - 4X^{\frac{1}{2}} + 6X^{\frac{-1}{2}})}{dX}$

$\large \frac{dY}{dX} = - 2X^{\frac{-1}{2}} - 3X^{\frac{-3}{2}}$

E. $\large T = -2R^{-1}+7R^{-3}-48+R^2$

Differentiating T wrt R we get

$\large \frac{dT}{dR} = \frac{d(-2R^{-1}+7R^{-3}-48+R^2)}{dR}$

$\large \frac{dT}{dR} = 2R^{-2} -21R^{-4} + 2R$

F. $\large Q = 25 - 3S^{2} +5S^{-3}$

Differentiating Q wrt S we get

$\large \frac{dQ}{dS}= \frac{d(25 - 3S^{2} +5S^{-3})}{dS}$

$\large \frac{dQ}{dS}= - 6S - 15S^{-4}$

G. $\large B = 3Q^3(2Q - 3Q^2)$

$\large \implies B = 6Q^4 - 9Q^5$

Now differentiating B wrt Q we get

$\large \frac{d B}{dQ} = \frac{d(6Q^4 - 9Q^5)}{dQ}$

$\large \frac{d B}{dQ} = 24Q^3 - 45Q^4$

H. $\large Z = 20P^{\frac{1}{4}}(30P^{\frac{2}{5}})$

$\LARGE \implies Z = 600P^{\frac{1}{4}+\frac{2}{5}}$

$\LARGE \implies Z = 600P^{\frac{13}{20}}$

Differentiating Z wrt P we get

$\LARGE \frac{dZ}{dP}= 600\frac{d(P^{\frac{13}{20}})}{dP}$

$\LARGE \frac{dZ}{dP}= 600\frac{13}{20} P^{\frac{13}{20}-1}$

$\LARGE \frac{dZ}{dP}= 30*13*P^{\frac{13-20}{20}}$

$\LARGE \frac{dZ}{dP}= 390*P^{\frac{-7}{20}}$

only in the case of G and H it was required to simplify the equation then differentiating it.

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