How do you find the derivative for (-7x^2+8)^8(3x^2+9)^10?

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = - 4 x \left(189 {x}^{2} + 132\right) {\left(- 7 {x}^{2} + 8\right)}^{7} {\left(3 {x}^{2} + 9\right)}^{9}$

Explanation:

Using product rule of differentiation as follows

$\setminus \frac{d}{\mathrm{dx}} {\left(- 7 {x}^{2} + 8\right)}^{8} {\left(3 {x}^{2} + 9\right)}^{10}$

$= {\left(- 7 {x}^{2} + 8\right)}^{8} \setminus \frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 9\right)}^{10} + {\left(3 {x}^{2} + 9\right)}^{10} \setminus \frac{d}{\mathrm{dx}} {\left(- 7 {x}^{2} + 8\right)}^{8}$

$= {\left(- 7 {x}^{2} + 8\right)}^{8} \left(10 {\left(3 {x}^{2} + 9\right)}^{9}\right) \setminus \frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 9\right) + {\left(3 {x}^{2} + 9\right)}^{10} \left(8 {\left(- 7 {x}^{2} + 8\right)}^{7}\right) \setminus \frac{d}{\mathrm{dx}} \left(- 7 {x}^{2} + 8\right)$

$= 10 {\left(- 7 {x}^{2} + 8\right)}^{8} {\left(3 {x}^{2} + 9\right)}^{9} \left(6 x\right) + 8 {\left(3 {x}^{2} + 9\right)}^{10} {\left(- 7 {x}^{2} + 8\right)}^{7} \setminus \left(- 14 x\right)$

$= 4 x {\left(- 7 {x}^{2} + 8\right)}^{7} {\left(3 {x}^{2} + 9\right)}^{9} \left(15 \left(- 7 {x}^{2} + 8\right) - 28 \left(3 {x}^{2} + 9\right)\right)$

$= 4 x {\left(- 7 {x}^{2} + 8\right)}^{7} {\left(3 {x}^{2} + 9\right)}^{9} \left(- 189 {x}^{2} - 132\right)$

$= - 4 x \left(189 {x}^{2} + 132\right) {\left(- 7 {x}^{2} + 8\right)}^{7} {\left(3 {x}^{2} + 9\right)}^{9}$