Find the derivative of (1) y = x + SiNx / X (2) y = cos2x / SiNx + cnsx

Find the derivative of (1) y = x + SiNx / X (2) y = cos2x / SiNx + cnsx

First question:
y′＝1＋（xcosx－sinx）/x^2.
Second question:
y′＝（－2sinxsin2x－cosxcos2x）/（sinx）^2－sinx.

Finding the derivative of y = cos2x-sin3x

Solution
y=cos2x-sin3x
y’=(cos2x-sin3x)'
=(cos2x)'-(sin3x)'
=-2sin2x-3cos3x

Find the derivative of the function y = (1 + cos2x) 2

y′=2（1+cos2x）（1+cos2x）′=2（1+cos2x）（-sin2x）（2x）′=4（1+cos2x）（-sin2x）=-4sin2x-2sin4x

Let y = 1 / 2 cos2x, then what is the derivative of Y
The product of 1 / 2 and cos2x

y'=(1/2)(-sin2x 2)=-sin2x

Derivative of y = ㏑ cos2x
There is 1-2sin ^ 2x = cos2x in the formula
According to this formula, (cos2x) '= - 2Sin ^ 2x
Why is (cos2x) '= - 2sin2x in the answer?

(cos2x)'=(-2sin^2x)'
=-2*(sin²x)'
=-2*2sinx*(sinx)'
=-2*(2sinxcosx)
=-2sin2x

Find the derivative of this y = cos2x + xtanx

y=cos2x+xtanx
y'=(cos2x)'+(xtanx)'
=-sin2x*(2x)'+tanx+x(tanx)'
=-2sin2x+tanx+xsec^2x.

Finding the derivative of y = sin (2x + 3)

The derivative of y = sin (e ^ x-1)

y'=[sin(e^x－1)]'
=[cos(e^x－1)]×[(e^x－1)']
=[cos(e^x－1)]×[e^x]
=(e^x)cos(e^x－1)

What is the derivative of y = (1 + SiN x)?

Y 'denotes the derivative of Y
y`=(1+sinx)'=1`+(sinx)`=cosx
The derivative of the constant is 0 and the derivative of SiNx is cosx

X-0, find the limit of limx ^ X
X-0, find the limit of limx ^ x, limx ^ x = Lim e ^ LNX ^ x = Lim e ^ x * LNX ^ = e ^ limx * LNX. What can Lim raise?

It should be x → 0+
E ^ X and LNX are continuous functions
See limit and continuity of composite function