How do you differentiate #y=(x+5)(2x-3)(3x^2+4)#?

Question

Luke Alvarez · Accepted Answer

#y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)#

#y'=24x^3+63x^2-74x+28# If #y=uvw#, where #u#, #v#, and #w# are all functions of #x#, then: #y'=uvw'+uv'w+u'vw# (This can be found by doing a chain rule with two functions substitued as one, i.e. making #uv=z#) #u=x+5# #u'=1# #v=2x-3# #v'=2# #w=3x^2+4# #w'=6x# #y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)# #y'=6x^3+8x-9x^2-12+6x^3+8x+30x^2+40+12x^3+60x^2-18x^2-90x# #y'=24x^3+63x^2-74x+28#

Luke Alvarez · Answer

#dy/dx=24x^3+63x^2-74x+28# #"expand the factors and differentiate using the "color(blue)"power rule"# #•color(white)(x)d/dx(ax^n)=nax^(n-1)# #y=(x+5)(2x-3)(3x^2+4)# #color(white)(y)=6x^4+21x^3-37x^2+28x-60# #rArrdy/dx=24x^3+63x^2-74x+28#

Isabella Ackerman · Answer

undefined To differentiate the function $ y = (x+5)(2x-3)(3x^2+4) $, you can use the product rule of differentiation:

$$ \frac{d}{dx}[f(x)g(x)h(x)] = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) $$

1. Find the derivatives of each term:
   - $ f(x) = x + 5 $ (first factor)
   - $ g(x) = 2x - 3 $ (second factor)
   - $ h(x) = 3x^2 + 4 $ (third factor)

2. Find the derivatives of each factor:
   - $ f'(x) = 1 $ (derivative of $ x + 5 $)
   - $ g'(x) = 2 $ (derivative of $ 2x - 3 $)
   - $ h'(x) = 6x $ (derivative of $ 3x^2 + 4 $)

3. Apply the product rule:
   $$ y' = (1)(2x-3)(3x^2+4) + (x+5)(2)(3x^2+4) + (x+5)(2x-3)(6x) $$

4. Simplify the expression to obtain the final derivative.