Physics and Mathematics
Differentiation of a Function with Respect to Another Function
1. Concept Overview
Sometimes we need to differentiate one function with respect to another function, instead of differentiating both with respect to the same variable.
If:
[ y = f(x) ]
[ u = g(x) ]
Then the derivative of ( y ) with respect to ( u ) is:
[\dfrac{dy}{du} = \dfrac{\dfrac{dy}{dx}}{\dfrac{du}{dx}}]
This simply means:
Differentiate both functions with respect to the same variable (usually (x)), then divide the derivatives.
2. Explanation Intuition
If both (y) and (u) are dependent on (x):
- How fast (y) changes per small change in (x)
- How fast (u) changes per small change in (x)
Then:
[\dfrac{dy}{du}] [= \text{(Change in } y \text{ per unit change in } u\text{)}]
3. Key Formula (Always remember)
[\dfrac{dy}{du}][ = \dfrac{dy/dx}{du/dx}]
provided that [\dfrac{du}{dx} \neq 0]
Example 1
Find ([ \dfrac{dy}{du} ]) if
[y = x^{3} + 5x]
[u = 2x^{2} – 3]
Solution:
Step 1: Differentiate with respect to [x]:
[
\dfrac{dy}{dx} = 3x^{2} + 5
]
[
\dfrac{du}{dx} = 4x
]
Step 2: Divide
[
\dfrac{dy}{du} = \dfrac{3x^{2} + 5}{4x}
]
✔ Final Answer:
[
\boxed{\dfrac{dy}{du} = \dfrac{3x^{2} + 5}{4x}}
]
Example 2
If
[y = \sin x]
[u = \cos x]
Step 1: Differentiate
[
\dfrac{dy}{dx} = \cos x
]
[
\dfrac{du}{dx} = -\sin x
]
Step 2: Divide
[
\dfrac{dy}{du} = \dfrac{\cos x}{-\sin x} = -\cot x
]
✔ Final Answer:
[
\boxed{\dfrac{dy}{du} = -\cot x}
]
Example 3
Find [\dfrac{dy}{du}] if
[y = e^{x}] and [u = \ln(x)]
Step 1: Differentiate both w.r.t. [x]
[\dfrac{dy}{dx} = e^{x}]
[\dfrac{du}{dx} = \dfrac{1}{x}]
Step 2: Divide
[\dfrac{dy}{du}][ = \dfrac{e^{x}}{1/x}][ = xe^{x}]
✔ Final Answer:
[\dfrac{dy}{du} = xe^{x}]
Example 4
Find [\dfrac{dy}{du}] if
[y = \tan x] and [u = x^{2}]
Step 1: Differentiate:
[\dfrac{dy}{dx} = \sec^{2}x]
[\dfrac{du}{dx} = 2x]
Step 2: Divide:
[\dfrac{dy}{du} = \dfrac{\sec^{2}x}{2x}]
✔ Final Answer:
[\dfrac{dy}{du} = \dfrac{\sec^{2}x}{2x}]
Example 5
Find [\dfrac{dy}{du}] if:
[y = x + \dfrac{1}{x}]
[u = x – \dfrac{1}{x}]
Step 1: Differentiate:
[\dfrac{dy}{dx} = 1 – \dfrac{1}{x^{2}}]
[\dfrac{du}{dx} = 1 + \dfrac{1}{x^{2}}]
Step 2: Divide:
[\dfrac{dy}{du} = \dfrac{1 – \dfrac{1}{x^{2}}}{,1 + \dfrac{1}{x^{2}},}]
✔ Final Answer:
[\dfrac{dy}{du}][ = \dfrac{x^{2} – 1}{x^{2} + 1}]
4. Conceptual Questions with Solutions
1. Why do we divide derivatives to get [\dfrac{dy}{du}]?
Because both [y] and [u] depend on [x]. Their rates of change are related through: [\dfrac{dy}{du}][ = \dfrac{dy/dx}{du/dx}] It expresses how fast [y] changes when [u] changes.
2. What condition must hold for [\dfrac{dy}{du}] to exist?
[\dfrac{du}{dx} \neq 0] must hold. If [\dfrac{du}{dx} = 0], the derivative becomes undefined.
3. If [y] is constant and [u] is changing, what is [\dfrac{dy}{du}]?
If [y] is constant ⇒ [\dfrac{dy}{dx} = 0] Thus [\dfrac{dy}{du} = 0]
4. If [u] is constant but [y] is changing, what happens?
[\dfrac{du}{dx} = 0] → division by zero → [\dfrac{dy}{du}] is undefined.
5. Can [\dfrac{dy}{du}] be negative?
Yes. If [y] increases while [u] decreases, the ratio becomes negative.
6. What if both [y] and [u] do not depend on [x]?
Both derivatives zero → indeterminate form \[\dfrac{0}{0}\]. [\dfrac{dy}{du}] cannot be determined.
7. Is the chain rule related to this?
Yes. [\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}] This is the chain rule rearranged.
8. Can [\dfrac{dy}{du}] be written as [\dfrac{dx}{du}] × [\dfrac{dy}{dx}]?
Yes. It is equivalent: [\dfrac{dy}{du} = \dfrac{dy}{dx} \cdot \dfrac{dx}{du}]
9. Can parametric differentiation be expressed similarly?
Yes, same rule applies when parameters are used.
10. Why is [u] often chosen as a simpler function?
To simplify the expression and make differentiation easier.
11. Does [\dfrac{dy}{du}] always give a simpler result?
Not always, but often useful for inverse relationships or substitution.
12. If [y = u], what is [\dfrac{dy}{du}]?
[\dfrac{dy}{du} = 1]
13. If [y = u^{2}], what is [\dfrac{dy}{du}]?
[\dfrac{dy}{du} = 2u]
14. Is this technique used in logarithmic differentiation?
Yes, because variables are expressed in terms of other variables.
15. Why do we need this concept in inverse trigonometry?
Inverse functions require differentiating one variable w.r.t. another related variable.
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