Find the derivative of y=x^2 / e^2x using the Quotient Rule, and simplify.
I really need help! please! :)
http://en.wikipedia.org/wiki/Quotient_rule
y'= (2x*e^2x-2e^2x*x^2)/e^4x
= (2x)(1-x)/e^2x
check my work, I did it in my head.
thank u so much! I checked it cause i used ur wikipedia link and understood it better. my teacher uses her own formula or something instead of d(x) or whatever. thank u soo much!! :)
Sure, I can help you with that! To find the derivative of y=x^2 / e^2x using the Quotient Rule, we first need to understand what the Quotient Rule is.
The Quotient Rule states that if we have a function of the form f(x) = g(x) / h(x), where both g(x) and h(x) are differentiable functions, then the derivative of f(x) can be found using the following formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / [h(x)]^2
In our case, g(x) = x^2 and h(x) = e^2x.
Now, let's differentiate each part individually.
To find g'(x), we can apply the Power Rule, which states that if we have a function of the form f(x) = x^n, then its derivative f'(x) = n*x^(n-1). So, in our case, g'(x) = 2x.
To find h'(x), we need to differentiate e^2x. The derivative of e^x is simply e^x. Since we have e^(2x), we use the Chain Rule. The Chain Rule states that if we have a composite function f(g(x)), then its derivative is f'(g(x)) * g'(x). In our case, f(u) = e^u and g(x) = 2x. So, f'(u) = e^u and g'(x) = 2. Therefore, h'(x) = f'(g(x)) * g'(x) = e^(2x) * 2 = 2e^(2x).
Now, we can substitute g'(x), h(x), and h'(x) into the Quotient Rule formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / [h(x)]^2
= (2x * e^2x - x^2 * 2e^(2x)) / [e^(2x)]^2
= (2xe^2x - 2x^2e^(2x)) / [e^(4x)]
To simplify the expression further, we can factor out a 2x from the numerator:
f'(x) = (2x(e^2x - xe^(2x))) / [e^(4x)]
So, the derivative of y=x^2 / e^2x using the Quotient Rule is (2x(e^2x - xe^(2x))) / [e^(4x)].
Hope this helps! Let me know if you have any further questions.