A higher number, derivative, when △ x → 0, △ y → 0, isn't it?

If differentiable, yes, when △ x → 0, △ y → 0 is FX continuous, if differentiable, it must be continuous

2 high number derivation exercises 1. Find the derivative dy / DX of the function y = f (x) determined by the following parametric equation (1)x=2t,y=t^2 (2)x=te^-t,y=e^t 2. Use the logarithmic derivation method to find the derivatives of the following functions (1)y=x^x Please ask for the detailed explanation of the above exercises, thank you~

1.(1)dy/dx=(dy/dt)/(dx/dt)=2t/2=t=x/2(2)dy/dx=(dy/dt)/(dx/dt)=e^t/(e^-t-te^-t)=y/(1/y-x)=y^2/(1-xy)2.y=x^x =e^(xlnx)lny=xlnxy'/y=lnx+1y'=y(lnx+1)=x^x(lnx+1)

Derivation of LN | cosx |

In a range, for example, when cosx > 0, it is equal to - SiNx / cosx = - TaNx
cosx

F (x) = ln (25 sin2x) derivation How to find the derivative of trigonometric function SiNx cosx TaNx and so on

F (x) '= 2cos2x / 25sin2xsinx' = cosxcosx '= - sinxtanx = SiNx / cosx, then the derivative is: TaNx' = (SiNx '* cosx SiNx * cosx') / (cosx) ^ 2 = 1 / (cosx) ^ 2 = (SiNx '* cosx SiNx * cosx') / (cosx) ^ 2 = 1 / (cosx) ^ 2

Derivative height y = ln (under x + root sign (1 + x ^ 2)) Don't copy and paste

Y = ln (x + root (1 + x ^ 2)) y '= 1 / (x + root (1 + x ^ 2)) * (x + root (1 + x ^ 2))' = 1 / (x + root (1 + x ^ 2)) * (1 + 1 / 2 * 2x / root (1 + x ^ 2)) = 1 / (x + root (1 + x ^ 2)) * (1 + X / root (1 + x ^ 2)) = 1 / (x + root (1 + x ^ 2)) * {[root (1 + x ^ 2) + x] / root (1 + x ^ 2)

Derivation of higher numbers. X = T ^ 2 + 2T y = ln (1 + T). Urgent If x = T ^ 2 + 2T y = ln (1 + T), then dy / DX =? The answer is 2 (e ^ (2t)) / (sect) ^ 2. I calculate it to be 1 / 2 (T + 1) ^ 2

Obviously you are right
Where did the answer come from? It's obviously wrong

A higher number, derivative, when △ x → 0, △ y → 0, isn't it?