# Question #7d7dc

Dec 7, 2017

You need to use implicit differentiation. Please see the explanation.

#### Explanation:

Given:

$P {V}^{\frac{7}{5}} = c$

Differentiate with respect to P:

$\frac{d \left(P {V}^{\frac{7}{5}}\right)}{\mathrm{dP}} = \frac{d \left(c\right)}{\mathrm{dP}}$

The term on the left is a product of two variables, therefore, we use the product rule:

$\frac{d \left(P\right)}{\mathrm{dP}} {V}^{\frac{7}{5}} + P \frac{d \left({V}^{\frac{7}{5}}\right)}{\mathrm{dP}} = \frac{d \left(c\right)}{\mathrm{dP}}$

The derivative of P with respect to P is 1:

${V}^{\frac{7}{5}} + P \frac{d \left({V}^{\frac{7}{5}}\right)}{\mathrm{dP}} = \frac{d \left(c\right)}{\mathrm{dP}}$

We use the chain rule on the derivative in the second term:

${V}^{\frac{7}{5}} + P \frac{d \left({V}^{\frac{7}{5}}\right)}{\mathrm{dV}} \frac{\mathrm{dV}}{\mathrm{dP}} = \frac{d \left(c\right)}{\mathrm{dP}}$

Because the derivative is with respect to V, we may apply the power rule:

${V}^{\frac{7}{5}} + \frac{7}{5} P {V}^{\frac{2}{5}} \frac{\mathrm{dV}}{\mathrm{dP}} = \frac{d \left(c\right)}{\mathrm{dP}}$

The derivative of the constant on the right is 0:

${V}^{\frac{7}{5}} + \frac{7}{5} P {V}^{\frac{2}{5}} \frac{\mathrm{dV}}{\mathrm{dP}} = 0$

Solve the above equation for $\frac{\mathrm{dV}}{\mathrm{dP}}$:

$\frac{7}{5} P {V}^{\frac{2}{5}} \frac{\mathrm{dV}}{\mathrm{dP}} = - {V}^{\frac{7}{5}}$

$\frac{\mathrm{dV}}{\mathrm{dP}} = - {V}^{\frac{7}{5}} / \left(\frac{7}{5} P {V}^{\frac{2}{5}}\right)$

$\frac{\mathrm{dV}}{\mathrm{dP}} = - \frac{5}{7} \frac{V}{P}$