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

Question

William Abdalla · Accepted Answer

#V'(x)=14x^6-4x^3-6# Applying the rule of product #(fg)'=f'g+fg'# we get #V'(x)=6x^2(x^4-2x)+(2x^3+3)(4x^3-2)# Growing, we attain #V'(x)=6x^6-12x^3+8x^6+12x^3-4x^3-6# #V'(x)=14x^6-4x^3-6#

Charles Arocha · Answer

#V'(x)=14x^6-4x^3-6# #"differentiate using the "color(blue)"power rule"# #•color(white)(x)d/dx(ax^n)=nax^(n-1)# #"expand the factors using FOIL"# #V(x)=2x^7-4x^4+3x^4-6x# #color(white)(V(x))=2x^7-x^4-6x# #V'(x)=14x^6-4x^3-6#

Charles Anctil · Answer

undefined To differentiate the function $ V(x) = (2x^3 + 3)(x^4 - 2x) $, you would use the product rule of differentiation. Here's how you apply it:

1. Identify the two functions being multiplied: $ f(x) = 2x^3 + 3 $ and $ g(x) = x^4 - 2x $.
2. Apply the product rule: $ V'(x) = f'(x)g(x) + f(x)g'(x) $.
3. Differentiate $ f(x) $ and $ g(x) $ separately.
4. Substitute the derivatives and the original functions into the product rule formula.
5. Simplify the expression to get the derivative of $ V(x) $.

Here's the step-by-step process:

1. $ f(x) = 2x^3 + 3 $ and $ g(x) = x^4 - 2x $.
2. $ f'(x) = 6x^2 $ and $ g'(x) = 4x^3 - 2 $.
3. Product rule: $ V'(x) = (6x^2)(x^4 - 2x) + (2x^3 + 3)(4x^3 - 2) $.
4. Expand and simplify: $ V'(x) = 6x^6 - 12x^3 + 8x^6 - 4x^3 + 12x^3 - 6 $.
5. Combine like terms: $ V'(x) = 14x^6 - 16x^3 - 6 $.

So, the derivative of $ V(x) $ is $ V'(x) = 14x^6 - 16x^3 - 6 $.