Question Statement
Differentiate the following expression with respect to x:
aβxa+xβ
Background and Explanation
To solve this problem, we will use the quotient rule for differentiation. The quotient rule is applied when you are differentiating a function that is the ratio of two functions, such as v(x)u(x)β. The quotient rule states:
dxdβ(v(x)u(x)β)=(v(x))2v(x)β
uβ²(x)βu(x)β
vβ²(x)β
Here, u(x)=a+x and v(x)=aβx, so we need to differentiate both the numerator and the denominator.
Solution
Let the function be:
Y=aβxa+xβ
Now, differentiate both sides with respect to x using the quotient rule:
-
Differentiate the numerator (u(x)=a+x):
uβ²(x)=1
-
Differentiate the denominator (v(x)=aβx):
vβ²(x)=β1
-
Apply the quotient rule:
Using the formula for the quotient rule:
dxdyβ=(v(x))2v(x)β
uβ²(x)βu(x)β
vβ²(x)β
Substitute the values for u(x), v(x), uβ²(x), and vβ²(x):
dxdyβ=(aβx)2(aβx)β
1β(a+x)β
(β1)β
-
Simplify the expression:
dxdyβ=(aβx)2(aβx)+(a+x)β
Combine like terms in the numerator:
dxdyβ=(aβx)22aβ
Thus, the derivative of the given function is:
dxdyβ=(aβx)22aβ
Final Answer:
dxdyβ=(aβx)22aβ
- Quotient Rule:
If you have a function in the form v(x)u(x)β, the derivative is:
dxdβ(v(x)u(x)β)=(v(x))2v(x)β
uβ²(x)βu(x)β
vβ²(x)β
Summary of Steps
-
Identify the function:
Y=aβxa+xβ
-
Apply the quotient rule:
- Differentiate the numerator u(x)=a+x to get uβ²(x)=1
- Differentiate the denominator v(x)=aβx to get vβ²(x)=β1
-
Substitute into the quotient rule formula:
dxdyβ=(aβx)2(aβx)β
1β(a+x)β
(β1)β
-
Simplify the expression:
dxdyβ=(aβx)22aβ
-
Final Answer:
dxdyβ=(aβx)22aβ