Find 𝑑𝑦𝑑𝑥 if 𝑥 = 3 + 𝑐𝑜𝑠𝑡 and 𝑦 = 1 − 𝑠𝑖𝑛𝑡

To find 𝑑𝑦𝑑𝑥, we need to differentiate the given equations with respect to 𝑥.

Given: 𝑥 = 3 + 𝑐𝑜𝑠𝑡 𝑦 = 1 − 𝑠𝑖𝑛𝑡

Step 1: Differentiate 𝑥 with respect to 𝑥 Since 𝑥 is a function of 𝑥 itself, the derivative of 𝑥 with respect to 𝑥 is 1.

So, 𝑑𝑥𝑑𝑥 = 1

Step 2: Differentiate 𝑦 with respect to 𝑥 To differentiate 𝑦 with respect to 𝑥, we need to use the chain rule.

Let's differentiate 𝑦 = 1 − 𝑠𝑖𝑛𝑡 with respect to 𝑥.

𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑥

The derivative of 𝑠𝑖𝑛𝑡 with respect to 𝑥 is given by: 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑥 = 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑡 * 𝑑𝑡𝑑𝑥

To find 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑡, we can use the derivative of the sine function, which is 𝑐𝑜𝑠𝑡.

So, 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑡 = 𝑐𝑜𝑠𝑡

To find 𝑑𝑡𝑑𝑥, we can differentiate 𝑥 = 3 + 𝑐𝑜𝑠𝑡 with respect to 𝑡 and then divide by 𝑑𝑥𝑑𝑡.

𝑑(𝑥)𝑑𝑡 = 𝑑(3 + 𝑐𝑜𝑠𝑡)𝑑𝑡 = 𝑑(3)𝑑𝑡 + 𝑑(𝑐𝑜𝑠𝑡)𝑑𝑡 = 0 + 𝑐𝑜𝑠𝑡 = 𝑐𝑜𝑠𝑡

𝑑𝑡𝑑𝑥 = 1 / 𝑑(𝑥)𝑑𝑡 = 1 / 𝑐𝑜𝑠𝑡

Now, we can substitute the values we found back into the chain rule equation:

𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑡 * 𝑑𝑡𝑑𝑥

Substituting the values we found: 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * 𝑑(𝑠𝑖𝑛𝑡)𝑑𝑡 * 𝑑𝑡𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * 𝑐𝑜𝑠𝑡 * (1 / 𝑐𝑜𝑠𝑡)

Simplifying the equation: 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) * (1 / 𝑐𝑜𝑠𝑡) 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) / 𝑐𝑜𝑠𝑡

Since 𝑦 = 1 − 𝑠𝑖𝑛𝑡, the derivative of 𝑦 with respect to 𝑠𝑖𝑛𝑡 is simply -1.

So, 𝑑𝑦𝑑(𝑠𝑖𝑛𝑡) = -1

Substituting this value back into the equation: 𝑑𝑦𝑑𝑥 = -1 / 𝑐𝑜𝑠𝑡

Therefore, 𝑑𝑦𝑑𝑥 = -1 / 𝑐𝑜𝑠𝑡.