# How do you differentiate f(x) + x^2 [f(x)]^3 = 30?

May 9, 2018

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = - \frac{2 x {\left[f \left(x\right)\right]}^{3}}{1 + 3 {x}^{2} {\left[f \left(x\right)\right]}^{2}}$

#### Explanation:

Let $y = f \left(x\right)$, then we can write $f \left(x\right) + {x}^{2} {\left[f \left(x\right)\right]}^{3} = 30$ as

$y + {x}^{2} {y}^{3} = 30$ and differentiating it implicitly

$\frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{3} + {x}^{2} \cdot 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left[1 + 3 {x}^{2} {y}^{2}\right] = - 2 x {y}^{3}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x {y}^{3}}{1 + 3 {x}^{2} {y}^{2}}$

or $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = - \frac{2 x {\left[f \left(x\right)\right]}^{3}}{1 + 3 {x}^{2} {\left[f \left(x\right)\right]}^{2}}$

May 9, 2018

The answer is $= - \frac{2 x {\left(f \left(x\right)\right)}^{3}}{1 + 3 {x}^{2} {\left(f \left(x\right)\right)}^{2}}$

#### Explanation:

Another method for implicit differentiation

The function is

$f \left(x\right) + {x}^{2} {\left(f \left(x\right)\right)}^{3} = 30$

Let $y = f \left(x\right)$

Then,

$y + {x}^{2} {y}^{3} - 30 = 0$

Let

$f \left(x , y\right) = y + {x}^{2} {y}^{3} - 30$

Then,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$\frac{\partial f}{\partial x} = 2 x {y}^{3}$

$\frac{\partial f}{\partial x} = 1 + 3 {y}^{2} {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x {y}^{3}}{1 + 3 {x}^{2} {y}^{2}} = - \frac{2 x {\left(f \left(x\right)\right)}^{3}}{1 + 3 {x}^{2} {\left(f \left(x\right)\right)}^{2}}$