How do you find the derivative of y=(3x^2+5)(7x^3+8x) ?

Aug 27, 2015

${y}^{'} = 105 {x}^{4} + 177 {x}^{3} + 40$

Explanation:

Use the distributive property of multiplication to rewrite your original function as

$y = 21 {x}^{5} + 24 {x}^{3} + 35 {x}^{3} + 40 x$

$y = 21 {x}^{5} + 59 {x}^{3} + 40 x$

You can now differentiate this function by using the power rule

$\frac{d}{\mathrm{dx}} \left(y\right) = 21 \cdot 5 {x}^{4} + 59 \cdot 3 {x}^{2} + 40$

${y}^{'} = \textcolor{g r e e n}{105 {x}^{4} + 177 {x}^{3} + 40}$

Altrnatively, you can actually go about differentiating this function by using the product rule as well

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

for

$f \left(x\right) = 3 {x}^{2} + 5 \text{ }$ and $\text{ } g \left(x\right) = 7 {x}^{3} + 8 x$

This will get you

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 5\right)\right] \cdot \left(7 {x}^{3} + 8 x\right) + \left(3 {x}^{2} + 5\right) \cdot \frac{d}{\mathrm{dx}} \left(7 {x}^{3} + 8 x\right)$

${y}^{'} = 6 x \cdot \left(7 {x}^{3} + 8 x\right) + \left(3 {x}^{2} + 5\right) \cdot \left(21 {x}^{2} + 8\right)$

${y}^{'} = 42 {x}^{4} + 48 {x}^{2} + 63 {x}^{4} + 24 {x}^{2} + 105 {x}^{2} + 40$

${y}^{'} = \textcolor{g r e e n}{105 {x}^{4} + 177 {x}^{2} + 40}$