# How do you differentiate  y = (2x^3 + 4 )/( x + 7)?

Oct 14, 2015

${f}^{'} \left(x\right) = \frac{4 {x}^{3} + 42 {x}^{2} + 4}{x + 7} ^ 2$

#### Explanation:

If we can notate the function in the form of

$f \left(x\right) = g \frac{x}{h \left(x\right)}$

We can say that

${f}^{'} \left(x\right) = \frac{{g}^{'} \left(x\right) h \left(x\right) - g \left(x\right) {h}^{'} \left(x\right)}{{h}^{2} \left(x\right)}$, so

${f}^{'} \left(x\right) = \frac{\left(6 {x}^{2}\right) \left(x + 7\right) - \left(2 {x}^{3} + 4\right) \left(1\right)}{x + 7} ^ 2$

${f}^{'} \left(x\right) = \frac{6 {x}^{3} + 42 {x}^{2} - 2 {x}^{3} + 4}{x + 7} ^ 2$

${f}^{'} \left(x\right) = \frac{4 {x}^{3} + 42 {x}^{2} + 4}{x + 7} ^ 2$