# 6(4x+3)^5(4)(3x-5)^5+ 5(3x-5)^4(3)(4X+3)^6

common factors: (4x+3)^5 (3x-5)^4

[(4x+3)^5 (3x-5)^4] [24(3x-5) + 15(4x+3)]

Take the second term, combine like terms.

f(x)=(4x+3)^6(3x-5)^5 Find the derivative.

This is what I did so far.
=6(4x+3)^5(4)(3x-5)^5+ 5(3x-5)^4(3)(4X+3)^6
=24(4X+3)^5(3x-5)^5 *[(3x-5)(15)(4x+3)]
=
I am not sure what to do next. Also did I do this correctly so far?

8 months ago

## You have done the first step correctly by factoring out the common factors of (4x+3)^5 and (3x-5)^4. Now, let's simplify the expression further.

Starting from where you left off:
= 24(4x+3)^5(3x-5)^5 * [(3x-5)(15)(4x+3)]

To simplify this expression, let's distribute the terms inside the brackets:
= 24(4x+3)^5(3x-5)^5 * (45x^2 - 225x - 75x + 375)

Next, combine like terms within the brackets:
= 24(4x+3)^5(3x-5)^5 * (45x^2 - 300x + 375)

Now we have simplified the expression, and we can see that we have a common factor of (4x+3)^5(3x-5)^5.
So, the final expression is:
= [(4x+3)^5(3x-5)^5] * (45x^2 - 300x + 375)

Now, let's move on to the second question you asked. To find the derivative of f(x)=(4x+3)^6(3x-5)^5, we can use the product rule of differentiation.

The product rule states that if we have two functions u(x) and v(x), then the derivative of their product u(x)v(x) is given by:
(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)

Let's apply the product rule to find the derivative of f(x):
f'(x) = [(4x+3)^6]'(3x-5)^5 + (4x+3)^6[(3x-5)^5]'

Now, to differentiate (4x+3)^6, we can apply the chain rule:
[(4x+3)^6]' = 6(4x+3)^5 * (4)' = 24(4x+3)^5

To differentiate (3x-5)^5, we can again apply the chain rule:
[(3x-5)^5]' = 5(3x-5)^4 * (3)' = 15(3x-5)^4

Substituting these results back into the expression, we have:
f'(x) = 24(4x+3)^5(3x-5)^5 + (4x+3)^6 * 15(3x-5)^4

So, the derivative of f(x) is:
f'(x) = 24(4x+3)^5(3x-5)^5 + 15(3x-5)^4(4x+3)^6

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