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

1 Answer
Aug 27, 2015

Answer:

#y^' = 105x^4 + 177x^3 + 40#

Explanation:

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

#y = 21x^5 + 24x^3 + 35x^3 + 40x#

#y = 21x^5 + 59x^3 + 40x#

You can now differentiate this function by using the power rule

#d/dx(y) = 21 * 5x^4 + 59 * 3x^2 + 40#

#y^' = color(green)(105x^4 + 177x^3 + 40)#

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

#color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))#

for

#f(x) = 3x^2 + 5" "# and #" "g(x) = 7x^3 + 8x#

This will get you

#d/dx(y) = [d/dx(3x^2 + 5)] * (7x^3 + 8x) + (3x^2 + 5) * d/dx(7x^3 + 8x)#

#y^' = 6x * (7x^3 + 8x) + (3x^2 + 5) * (21x^2 + 8)#

#y^' = 42x^4 + 48x^2 + 63x^4 + 24x^2 + 105x^2 + 40#

#y^' = color(green)(105x^4 + 177x^2 + 40)#