If y = the square of 2x + INX, find the square of the second derivative y of Y

If y = the square of 2x + INX, find the square of the second derivative y of Y

Y=2x^2+Inx
y′=4x+1/x
y″=4-1/x^2
I don't understand the back

Mathematics for Liberal Arts: finding the derivative of F (x) = [cos (INX)] ^ 2
There should be a process. Thank you

f(x)= [ cos(Inx) ] ^2
f'(x)=2cos(lnx)*(cos(lnx))'
=2cos(lnx)*(-sin(lnx))*(lnx)'
=-2cos(lnx)*sin(lnx)*1/x
=-[sin(2lnx)]/x

X = a (t-sint), y = a (1-cost), please construct the binary function f (x, y) about X, y, such that f (x, y)

4a[1-cos(t/2)]=8a[sin(t/4)]^21-cost=2[sin(t/2)]^2sint=2sin(t/2)cos(t/2)tan(t/2)=(1-cost)/sintcot(t/2)=sint/(1-cost)[sin(t/2)]^2=1/[1+[cot(t/2)]^2]=(1-cost)^2/[(sint)^2+(1-cost)^2](1-cost)^2/[[(t-sint)...

Let z = u ^ V, u = e ^ x, v = under the root sign (x ^ 2 + 1), find DZ / DX

To find DZ / DX is to find the total derivative
According to the chain rule:
DZ / DX = (partial Z / partial U) · Du / DX + (partial Z / partial V) · DV / DX
=v·u^(v-1)·e^x+u^v·lnu·[1/2√(x²+1)]·2x
=v·u^(v-1)·e^x+u^v·lnu·x/√(x²+1)
Then u = e ^ x, v = √ (x ^ 2 + 1) is substituted to get:
dz/dx=v·u^v+u^v·x·x/v
=u^v·(v+x²/v)
=[(e^x)^√(x²+1)]·(2x²+1)/√(x²+1)
PS: (e ^ x) ^ √ (X & sup2; + 1) refers to the power of √ (X & sup2; + 1) with e ^ x as the base
Partial differential symbol can not be typed, please understand!

Let z = arctan (UV), u = e ^ x, v = x ^ 3, find DZ / DX

Tan (z) = UV = x ^ 3 e ^ (x) dtan (z) / DX = sec ^ 2 (z) DZ / DX = 3x ^ 2E ^ (x) + x ^ 3E ^ (x) the solution is: DZ / DX = x ^ 2 (3 + x) e ^ (x) / sec ^ 2 (z) / /: 1 + Tan ^ 2 (z) = sec ^ 2 (z) = x ^ 2 (x + 3) e ^ (x) / [1 + x ^ 6 e ^ (2x)]

Z = u & # 178; V + 3UV ^ 4, u = e ^ x, v = SiN x, find DZ / DX, ask the gods to answer →_ →

Z = u & # 178; V + 3UV ^ 4, u = e ^ x, v = SiN x, find DZ / DX
dz/dx=2uu'v+u^2v'+3u'v^4+3v(4v^3)v'
=2e^(2x)sinx + e^(2x)cosx +3e^x (sinx)^4 + 12(sinx)^4 cosx
=2sinx e^(2x) + cosx e^(2x) + 3(sinx)^4 e^x (e^x + 4cosx)
=2sinx e^(2x) + 3(sinx)^4 e^(2x) + cosx e^(2x)+12(sinx)^4 cosx e^(x)
=sinx e^(2x)[2+3(sinx)^3] + cosx e^x [e^x+12(sinx)^4]

Find the second derivative of the function determined by the following parametric equation: x = a (t-sint), y = a (1-cost)

dx/dt=a(1-cost)dy/dt=asinty'=sint/(1-cost)dy'/dt=[cost(1-cost)-sint*sint]/(1-cost)^2=(cost-1)/(1-cost)^2=-1/(1-cost)y"=(dy'/dt)/(dx/dt)=-1/[a(1-cost)^2]

Parameter equation x = t (1-cost) and y = TSINT determine the derivative of the function to find the answer! Thank you

dy/dx = y'(t) / x'(t) = ( sint + t cost) / ( 1- cost + t sint)

Find the second derivative of the function determined by x = e ^ t * cost, y = e ^ t * Sint
x't=(e^t)(sint+cost)
y't=(e^t)(cost-sint)
x''t=(e^t)(sint+cost+cost-sint)=2(e^t)cost
y''t=(e^t)(cost-sint-sint+cost)=-(e^t)sint
dy/dx=(cost-sint)/(sint+cost)
d^2 y/d(x^2)
=d(dy/dx)/dx
=(y''x'-y'x'')/(x')^2
What do X't and Y't mean?

1. It means that x is the dependent variable and t is the independent variable. It is a derivative of the function!

Find the derivative of the parametric function determined by the equation x = a (T + Sint), y = a (1 + cost). A is a constant

Dy / dt = - asint DX / dt = a + acost dy / DX = dy / DT / DX / dt = - asint / A + acost = - Sint / 1 + cost the result is independent of A