How do you differentiate #y = 1/2 x + 1/4sin2x#?

Question

Cameron Alexander · Accepted Answer

#dy/dx = cos^2x#

This can be rewritten as follows using the identity #sin2theta = 2sinthetacostheta# and a little algebra.

#y = 1/2x + 1/4(2sinxcosx)#

#y = 1/2x + 1/2sinxcosx#

#y = 1/2(x + sinxcosx)#

#2y = x + sinxcosx#

Use implicit differentiation and the product rule to differentiate.

#2(dy/dx) = 1 + cosx(cosx) + sinx(-sinx)#

#2(dy/dx) = 1 + cos^2x - sin^2x#

Use the identity #cos^2theta + sin^2theta = 1 ->1 - sin^2theta = cos^2theta#:

#2(dy/dx) = cos^2x + cos^2x#

#2(dy/dx) = 2cos^2x#

#dy/dx = (2cos^2x)/2#

#dy/dx= cos^2x#

Hopefully this helps!

Emilia Aguilar · Answer

undefined To differentiate $ y = \frac{1}{2}x + \frac{1}{4}\sin(2x) $, you can use the rules of differentiation:

1. Differentiate each term separately.
2. For $ \frac{1}{2}x $, the derivative of $ x $ with respect to $ x $ is $ 1/2 $.
3. For $ \frac{1}{4}\sin(2x) $, apply the chain rule. The derivative of $ \sin(2x) $ with respect to $ x $ is $ 2\cos(2x) $.
4. Multiply the result of step 3 by the derivative of the inner function, which is $ 2 $.

So, the derivative of $ y $ with respect to $ x $ is:

$$ y' = \frac{1}{2} + \frac{1}{4} \times 2 \cos(2x) $$

Simplify:

$$ y' = \frac{1}{2} + \frac{1}{2} \cos(2x) $$

Thus, the derivative of $ y $ with respect to $ x $ is $ \frac{1}{2} + \frac{1}{2} \cos(2x) $.