Calc III: Finding Partial Derivatives for a Challenging Integral Function

[falcon0311]

From what I can remember, this wasn't covered in my class, however it was a bonus question on a test, and I was wondering if someone could take the time to enlighten me on this one.

Find $$\frac{\delta f }{ \delta x}$$ and $$\frac{\delta f }{ \delta y}$$ when

$$f(x,y) = \int _x ^{x^3y} \sin (t^2) dt$$

 

[Tide]

This may help:

$$\frac {d}{dx} \int^x f(x') dx' = f(x)$$

 

[M98Ranger]

Finding partial derivatives for an integral function can be a challenging task, but it is an important skill to have in Calculus III. To find the partial derivatives \frac{\delta f }{ \delta x} and \frac{\delta f }{ \delta y}, we will use the fundamental theorem of calculus and the chain rule.

First, let's rewrite the function as follows:

f(x,y) = \int _0 ^x \sin (t^2) dt + \int _x ^{x^3y} \sin (t^2) dt

Using the fundamental theorem of calculus, we can rewrite the first integral as:

\int _0 ^x \sin (t^2) dt = \frac{d}{dx} \int _0 ^x \sin (t^2) dt = \sin (x^2)

Next, we can use the chain rule to find the derivative of the second integral:

\frac{d}{dx} \int _x ^{x^3y} \sin (t^2) dt = \sin (x^6y^2) \cdot \frac{d}{dx} (x^3y) = 3x^2y \cdot \sin (x^6y^2)

Therefore, the partial derivative with respect to x is:

\frac{\delta f }{ \delta x} = \frac{d}{dx} ( \sin (x^2) + 3x^2y \cdot \sin (x^6y^2) ) = 2x \cos (x^2) + 6x^2y \cos (x^6y^2)

Similarly, we can find the partial derivative with respect to y using the chain rule:

\frac{\delta f }{ \delta y} = \frac{d}{dy} ( \sin (x^2) + 3x^2y \cdot \sin (x^6y^2) ) = 3x^2 \sin (x^6y^2)

In summary, to find the partial derivatives of a challenging integral function, we can use the fundamental theorem of calculus and the chain rule. It may seem daunting at first, but with practice, it will become easier to tackle such problems. I hope this explanation has helped you understand how to find partial derivatives for an integral function.