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Problem 11

# Find the first partial derivatives of the function. $$f(s, t)=\left(s^{2}-s t+t^{2}\right)^{3}$$

Expert verified
The first partial derivatives of the function $$f(s, t) = (s^2 - st + t^2)^3$$ are: $$\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$$ $$\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$$
See the step by step solution

## Step 1: Write down the given function with its composite function.

The given function is $$f(s, t) = (s^2 - st + t^2)^3$$. We can notice that this function is a composite of two functions. We can rewrite it as: $$f(s, t) = g(h(s, t))$$, where $$g(u) = u^3$$ and $$h(s, t) = s^2 - st + t^2$$.

## Step 2: Evaluate the derivatives of the inner and outer functions with respect to the given variables.

Now we need to find the derivatives of the inner function $$h(s, t)$$ and the outer function $$g(u)$$ with respect to our variables, $$s$$ and $$t$$. Let's first find the derivatives with respect to $$s$$: $$\frac{\partial h}{\partial s} = 2s - t$$ $$\frac{\partial g}{\partial u} = 3u^2$$ Now, let's find the derivatives with respect to $$t$$: $$\frac{\partial h}{\partial t} = -s + 2t$$ $$\frac{\partial g}{\partial u} = 3u^2$$ (the derivative of $$g$$ with respect to $$u$$ remains the same)

## Step 3: Apply the chain rule to find the first partial derivatives of the given function.

Now that we have our inner and outer function derivatives, we can apply the chain rule to find our partial derivatives of the given function $$f(s, t)$$. First, the partial derivative with respect to $$s$$: $$\frac{\partial f}{\partial s} = \frac{\partial g}{\partial h} \cdot \frac{\partial h}{\partial s}$$ Substitute the found derivatives: $$\frac{\partial f}{\partial s} = (3u^2) \cdot (2s - t)$$ Now, recall that $$u = h(s, t) = s^2 - st + t^2$$. Substitute this back into our equation: $$\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$$ Next, the partial derivative with respect to $$t$$: $$\frac{\partial f}{\partial t} = \frac{\partial g}{\partial h} \cdot \frac{\partial h}{\partial t}$$ Substitute the found derivatives: $$\frac{\partial f}{\partial t} = (3u^2) \cdot (-s + 2t)$$ Again, substitute $$u = h(s, t) = s^2 - st + t^2$$ back into our equation: $$\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$$

## Step 4: Write down the final first partial derivatives of the given function.

We found the first partial derivatives of the given function $$f(s, t) = (s^2 - st + t^2)^3$$: $$\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$$ $$\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$$

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