How do you differentiate #sin^2(3-x)#?

Question

Ezekiel Alexander · Accepted Answer

#2sin(x-3)cos(x-3)#

We use the chain rule here

#d/dx[f(x)]^n=[f(x)]^(n-1)f'(x)#

We also use these facts about #sin# and #cos#

#sin(-x)=-sinx#

#cos(-x)=cosx#

So

#d/dxsin^2(3-x)=-2sin(3-x)cos(3-x)=-2(-sin(x-3))cos(x-3)=2sin(x-3)cos(x-3)#

Charles Anctil · Answer

undefined To differentiate $ \sin^2(3 - x) $:

1. Recognize that $ \sin^2(3 - x) $ represents the square of the sine function of the expression $ (3 - x) $.

2. Apply the chain rule, treating $ 3 - x $ as the inner function.

3. Differentiate the outer function $ \sin^2(u) $ with respect to its inner function $ u $, and then multiply by the derivative of the inner function.

4. The derivative of $ \sin^2(u) $ with respect to $ u $ is $ 2\sin(u)\cos(u) $.

5. The derivative of $ 3 - x $ with respect to $ x $ is $ -1 $.

6. Substitute $ u = 3 - x $ into the derivative formula.

7. Differentiate the inner function $ u = 3 - x $ to get $ du/dx = -1 $.

8. Combine the results from steps 4 and 7 to get the final derivative.

Therefore, the derivative of $ \sin^2(3 - x) $ with respect to $ x $ is $ -2\sin(3 - x)\cos(3 - x) $.

Axel Alvarez · Answer

undefined To differentiate sin^2(3-x), you apply the chain rule and the power rule. The derivative is -2sin(3-x)cos(3-x).