partial derivative of sin^-1 (x/y) with respect to x
Answers
Answer:
When we first considered what the derivative of a vector function might mean, there was really not much difficulty in understanding either how such a thing might be computed or what it might measure. In the case of functions of two variables, things are a bit harder to understand. If we think of a function of two variables in terms of its graph, a surface, there is a more-or-less obvious derivative-like question we might ask, namely, how "steep'' is the surface. But it's not clear that this has a simple answer, nor how we might proceed. We will start with what seem to be very small steps toward the goal; surprisingly, it turns out that these simple ideas hold the keys to a more general understanding.
Step-by-step explanation:
Answer:
The partial derivative of sin⁻¹(x/y) with respect to x is:
∂f/∂x = (1/√(1 - (x/y)²)) * (1/y)
Step-by-step explanation:
We can solve this problem using the chain rule of differentiation. Let f(x,y) = sin⁻¹(x/y). Then, the partial derivative of f with respect to x can be computed as follows:
∂f/∂x = (∂f/∂u) * (∂u/∂x)
where u = x/y.
We can use the formula for the derivative of the inverse sine function:
d/dx sin⁻¹(u) = 1/√(1 - u²) * du/dx
Applying this formula to our function f, we get:
∂f/∂u = 1/√(1 - u²)
∂u/∂x = 1/y
Substituting these values, we obtain:
∂f/∂x = (∂f/∂u) * (∂u/∂x) = (1/√(1 - (x/y)²)) * (1/y)
Therefore, the partial derivative of sin⁻¹(x/y) with respect to x is:
∂f/∂x = (1/√(1 - (x/y)²)) * (1/y)
To learn more about derivatives: https://brainly.in/question/23505216
To learn more about functions: https://brainly.in/question/3961483
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