Hi,
We want to find the second derivative of the function y = x(2x+3)^5. To do this, we need to differentiate the expression twice with respect to x.
First, we apply the product rule to differentiate y with respect to x. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:
(uv)' = u'v + uv'
where u' and v' denote the first derivatives of u and v, respectively.
In this case, we can let u(x) = x and v(x) = (2x+3)^5. Then, using the product rule, we have:
y' = u'v + uv' = (1)(2x+3)^5 + x(5(2x+3)^4(2)) // u' = 1, v' = 5(2x+3)^4(2) = 10(2x+3)^4 = (2x+3)^4(2(2x+3) + 5x) // simplify the terms = (2x+3)^4(9x + 6)
Now we need to find the second derivative, so we'll apply the product rule again. This time we'll differentiate y' with respect to x, using the product rule:
y'' = (u'v)' + (uv')' = (u''v + u'v') + (u'v' + uv'') = u''v + 2u'v' + uv''
where u'' and v'' denote the second derivatives of u and v, respectively.
We know from above that u(x) = x and v(x) = (2x+3)^5, so we can find the first derivatives u' and v', which are:
u' = 1 v' = 5(2x+3)^4(2) = 10(2x+3)^4
To find u'' and v'', we need to differentiate u' and v' with respect to x:
u'' = 0 // since u' = 1 is a constant function, its derivative is 0 v'' = 10(2x+3)^4(2)(2) = 80(2x+3)^4
Now we can substitute these values into the formula for y'' that we derived above:
y'' = u''v + 2u'v' + uv'' = 0(2x+3)^5 + 2(2x+3)^4(10x) + x(80(2x+3)^4) = 20(2x+3)^4 + 80x(2x+3)^3
So the second derivative of y = x(2x+3)^5 is y'' = 20(2x+3)^4 + 80x(2x+3)^3
