What is the derivative of y = x ^ (SiNx)? What is the derivative of y = x ^ (SiNx)? y=x^(sinx) LNY = sinxlnx why?

What is the derivative of y = x ^ (SiNx)? What is the derivative of y = x ^ (SiNx)? y=x^(sinx) LNY = sinxlnx why?

For y = x ^ SiNx, take the natural logarithm on both sides to obtain
lny＝lnx^sinx→lny＝sinxlnx.
Then find the derivative on both sides, and you get
(1/y)·y'＝(sinx)'·lnx+sinx·(lnx)'
→(1/y)·y'＝cosxlnx+sinx·(1/x)
→y'＝y[cosxlnx+(1/x)sinx],
That is, y '= x ^ (SiNx) · [cosxlnx + (1 / x) SiNx]

Find the value range of function y = x-x3, X ∈ [0,2]

∵y′=1-3x2，x∈[0，2]，
Let y ′ > 0, the solution is: 0 ≤ x ＜
three
3，
Let y ′ < 0, the solution is:
three
3＜x≤2，
The function is in [0,
three
3) Increment, in（
three
3, 2] decreasing,
∴x=
three
3, y Max: 2
three
9，
When x = 0, y = 0, when x = 2, y = - 6,
The value range of the function is: [- 6, 2]
three
9]．

Derivative of y = (x ^ 3-1) / SiNx

y=(x^3-1)/sinx
y=（(x^3-1)/sinx）‘
=[(x ³- 1)'sinx-(x ³- 1)(sinx)']/(sinx) ²
=[3x ² sinx-(x ³- 1)cosx]/(sinx) ²

Finding the value range of y = x + 1-e ^ x on [- 1,2] by derivative method

y=x+1-e^x
y'=1-e^x
x> 0, y '< 0, the function is a subtractive function
X < 0, y '> 0, the function is an increasing function
Therefore, the maximum value is f (0) = 0 + 1-e ^ 0 = 0
The minimum value is the minimum of F (- 1) and f (2)
f(-1)=-1+1-e^(-1)=-1/e
f(2)=2+1-e^2=3-e ²<- 1<-1/e
Therefore, the minimum value is 3-e ²
So the range is [3-e ², 0]

How to find the derivative of y = xtanx?

(xtanx)'
=x'tanx+x(tanx)'
=tanx+x(sinx/cosx)'
=tanx+x[(sinx)'cosx-sinx(cosx)']/(cosx)^2
=tanx+x[1+(tanx)^2]
=tanx+x/(cosx)^2
=tanx+x*(secx)^2

Derivative of y = xtanx?

Y '= TaNx + xsec squared x