Question

Apply chain rule to calculate da/dx where a(x,y)= sin(xy).e×

Answer 1

Answer:

The chain rule states that the derivative of f(g(x)) is f'(g(x))⋅g'(x). ... For example, sin(x²) is a composite function because it can be constructed as f(g(x)) for f(x)=sin(x) and g(x)=x². Using the chain rule and the derivatives of sin(x) and x², we can then find the derivative of sin(x²)

Explanation:

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Answer 2

Solution:

Given => a(x,y) = sin(xy).eˣ

To find => $\frac{da}{dx}$

$\frac{da}{dx}$

$\frac{d}{dx} (sin(xy).e^{x})$

As we are differentiating with respect to x then y will be a constant.

Apply the Product Rule => (f.g)' = f'.g + f.g'

Here, f = sin(xy) and g = eˣ

$\frac{d}{dx} (sin(xy)).e^{x} + \frac{d}{dx} (e^{x}).sin(xy)$

Now, first let's solve $\frac{d}{dx} (sin(xy))$

$\frac{d}{dx} (sin(xy))$

Apply the Chain Rule => $\frac{df(u)}{dx} = \frac{df}{du} . \frac{du}{dx}$

Here, u = xy and f = sin(u)

$\frac{d}{du}(sin(u)) . \frac{d}{dx}(xy)$

$cos(u).y$

$cos(xy).y$

Now, let's solve $\frac{d}{dx} (e^{x})$

$\frac{d}{dx} (e^{x}) = e^{x}$

Now, putting the values back in this $\frac{d}{dx} (sin(xy)).e^{x} + \frac{d}{dx} (e^{x}).sin(xy)$$\frac{d}{dx} (sin(xy)).e^{x} + \frac{d}{dx} (e^{x}).sin(xy)$

$cos(xy)ye^{x} + e^{x}sin(xy)$

The $\frac{da}{dx}$ of a(x,y) = sin(xy).eˣ is $cos(xy)ye^{x} + e^{x}sin(xy)$