find the first and second derivatives of
(Ax^2+Bx+c)e^(3x)
should i distribute e^3x into the prenthesis? or is there an easy way....
b/c when i distribute it i get a long answer.....
for
i get
yp'=2Axe^3x+e^3xAx^2+Be^3x+e^3xBx+e^3x+ce^3x
is that correct...please help me....
thanks for ALLL YOUR generous help :)
You could distribute it or use the product rule. The result should be the same. It looks like you made a mistake differentiating e^3x. The answer should be 3 e^3x
o i see...thats correct...thanks so much...
To find the first and second derivatives of (Ax^2+Bx+c)e^(3x), you can use the product rule and the chain rule.
Let's start by finding the first derivative. To do this, we will apply the product rule. The product rule states that if you have two functions u(x) and v(x), the derivative of their product u(x)v(x) can be found using the formula:
(uv)' = u'v + uv'
In our case, u(x) = Ax^2 + Bx + c and v(x) = e^(3x).
First, we need to find the derivative of u(x), which is u'(x). The derivative of Ax^2 + Bx + c with respect to x is 2Ax + B.
Next, we need to find the derivative of v(x), which is v'(x). The derivative of e^(3x) with respect to x is 3e^(3x).
Now, we can apply the product rule to find the first derivative:
(uxv)' = u'v + uv'
So, applying the product rule, the first derivative becomes:
(2Ax + B)(e^(3x)) + (Ax^2 + Bx + c)(3e^(3x))
Simplifying further, we get:
2Axe^(3x) + Be^(3x) + Ax^2(3e^(3x)) + Bx(3e^(3x)) + 3ce^(3x)
This is the correct expression for the first derivative.
Now, to find the second derivative, we need to find the derivative of the expression we just obtained. We can use the product rule again.
Applying the product rule, the second derivative becomes:
(2Ax + B)(3e^(3x)) + (3e^(3x))(2Ax + B) + 2Ax(3e^(3x)) + B(3e^(3x))
Simplifying further, we get:
6Axe^(3x) + 3Be^(3x) + 6Axe^(3x) + 3Bx(e^(3x)) + 2Ax(3e^(3x)) + 3B(3e^(3x))
Combining like terms, we have:
12Axe^(3x) + 6Axe^(3x) + 6Axe^(3x) + 3Be^(3x) + 3Bx(e^(3x)) + 9Be^(3x) + 6Ax(e^(3x))
Simplifying further, we get:
24Axe^(3x) + 9Be^(3x) + 3Bx(e^(3x)) + 6Ax(e^(3x))
This is the correct expression for the second derivative.
Remember that when doing these calculations, it's important to be careful with signs and algebraic manipulations. It can be a bit time-consuming, but this method ensures that you get the correct derivatives.