find the derivate of the function Y=(2x+3)^4(3x+1)^-2?
To find the derivative of the function y = (2x+3)^4 * (3x+1)^(-2), we will use the product rule and the chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by:
(d/dx) [u(x) * v(x)] = u(x) * v'(x) + v(x) * u'(x)
To begin, let's find the derivative of the first term: (2x+3)^4.
We can rewrite this term as u(x) = (2x+3)^4. To find u'(x), we use the chain rule, which states that the derivative of the composition of two functions is the derivative of the outer function multiplied by the derivative of the inner function.
Let's define another function f(x) = 2x + 3. Taking the derivative of f(x) with respect to x, we get:
f'(x) = 2
Now, using the chain rule, the derivative of u(x) = (2x+3)^4 becomes:
u'(x) = 4 * (2x+3)^3 * (d/dx) [2x + 3] = 4 * (2x+3)^3 * 2
Next, let's find the derivative of the second term: (3x+1)^(-2).
We can rewrite this term as v(x) = (3x+1)^(-2). Again, we will use the chain rule.
Let's define another function g(x) = 3x + 1. Taking the derivative of g(x) with respect to x, we get:
g'(x) = 3
Using the chain rule, the derivative of v(x) = (3x+1)^(-2) becomes:
v'(x) = -2 * (3x+1)^(-3) * (d/dx) [3x + 1] = -2 * (3x+1)^(-3) * 3
Now, we can apply the product rule to find the derivative of the entire function y = (2x+3)^4 * (3x+1)^(-2).
(d/dx) [y] = (2x+3)^4 * v'(x) + (3x+1)^(-2) * u'(x)
= (2x+3)^4 * [-2 * (3x+1)^(-3) * 3] + (3x+1)^(-2) * [4 * (2x+3)^3 * 2]
= -6(2x+3)^4(3x+1)^(-3) + 8(2x+3)^3(3x+1)^(-2)
So, the derivative of the function y = (2x+3)^4 * (3x+1)^(-2) is:
y' = -6(2x+3)^4(3x+1)^(-3) + 8(2x+3)^3(3x+1)^(-2)