hi, just wondering on how i should approach differentiating: loge1/(3-x)^5
can i differentiate it as loge(3-x)^5?
sorry i meant to write (3-x)^-5
Hmmmm.
You are haveing a brain seizure. Watch this:
Ln x^-a= - ln x^a= -a ln x
Amazing.
So loge1/(3-x)^5= -5 ln(3-x)
Check my thinking.
To differentiate the function loge(1/(3-x)^5), you can apply the chain rule.
The chain rule states that if you have a composite function, f(g(x)), then the derivative of f(g(x)) with respect to x is equal to the derivative of f(g(x)) multiplied by the derivative of g(x) with respect to x.
In this case, the outer function is loge(u), where u = 1/(3-x)^5, and the inner function is u = 1/(3-x)^5.
To differentiate loge(u), you can use the chain rule as follows:
1. Identify the outer function: f(u) = loge(u)
2. Identify the inner function: g(x) = 1/(3-x)^5
Now, differentiate the outer function and the inner function:
1. Derivative of the outer function: df/du = 1/u
2. Derivative of the inner function: dg/dx = -5/(3-x)^(5-1) = -5/(3-x)^4
Next, apply the chain rule:
d/dx [f(g(x))] = df/du * dg/dx
= (1/u) * (-5/(3-x)^4)
= -5/(u * (3-x)^4)
Substituting u = 1/(3-x)^5:
= -5/(1/(3-x)^5 * (3-x)^4)
Simplifying:
= -5(3-x)^(-5) * (3-x)^4
= -5/(3-x)
Therefore, the derivative of loge(1/(3-x)^5) is -5/(3-x).