Find the derivative of implicit function y = 7SiN (π x + y)

Find the derivative of implicit function y = 7SiN (π x + y)

y'=7cos(πx+y)*(π+y')
You can solve y '

Derivative of implicit function e ^ y = x + y

There are two sides deriving X
y'e^y=1+y'
There are, y '= 1 / (e ^ Y-1)

Find the derivative of the implicit function x Λ y = y Λ X,

Take the natural logarithm on both sides
ylnx=xlny
Two side derivative
y'lnx+y/x=lny+xy'/y
Just figure out y '

Derivative of implicit function xlny(x)+y(x)e^(xy(x))-2=0 Finding y '(x) is the derivation of X

x(lny(x))'+lny(x)+y(x)(e^(xy(x)))'+y'(x)e^(xy(x))=0
x(1/y(x))y'(x)+lny(x)+y(x)(e^(xy(x)))(xy(x))'+y'(x)e^(xy(x))=0
x(1/y(x))y'(x)+lny(x)+y(x)(e^(xy(x)))(y(x)+xy'(x))+y'(x)e^(xy(x))=0

Y = logax derivation

Derivation rule from composite function
y'=1/(x*ln a)
a^y=x
Derivation of X on both sides:
y'*ln a*a^y=1
y'=1/(a^y*ln a)=1/(x*ln a)

F (x) = logax (x > 0), calculate the derivative

1/(xlna)
According to the bottom changing formula
f(x)=logax=lna/lnx
The law of derivation
f(x)=l/(xlna)

The derivative of the function f (x) defined on R at x = 0 is f '(0) = 1. Find the value of LIM f (2x) - f (- 3x) / X The derivative of the function f (x) defined on R at x = 0 is f '(0) = 1, Find the value of LIM f (2x) - f (- 3x) / X

lim (x→0)[ f(2x)-f(-3x)/x]
=lim (x→0)[ f(2x)-f(0)+f(0)-f(-3x)/x]
=lim (x→0){[f(2x)-f(0)]/x+[f(0)-f(-3x)]/x}
=lim (x→0)[ f(2x)-f(0)]/x- lim (x→0)[f(-3x)-f(0)]/x
=lim (x→0)2·[ f(2x)-f(0)]/2x- lim (x→0)(-3)·[f(-3x)-f(0)]/(-3x)
=2lim (x→0)[ f(2x)-f(0)]/2x+3 lim (x→0)[f(-3x)-f(0)]/(-3x)
=2f'(0)+3f'(0)=5f'(0)=5
=lim f(2x)-f(-3x)/lim f(2x)-f(-3x)/

F (x) = 2 / 3x ^ 2 (x1), the left and right derivatives of this function are calculated by definition, mainly the right derivative f(x)=2/3（x^2） I don't know how to get the right derivative by definition

If you don't say left derivative, you can get it;
Right derivative
=LIM (x tends to 1 from > 1) [f (x) - f (1)] / (x-1)
=LIM (x approaches 1 from > 1) [x ^ 2 - 2 / 3] / (x-1)
=LIM (x approaches 1 from > 1) [x ^ 2 - 1 + 1 / 3] / (x-1)
=LIM (x approaches 1 from > 1) (x ^ 2 - 1) / (x-1) + 1 / 3 LIM (x approaches 1 from > 1) 1 / (x-1)
The second function limit does not exist (the first limit is 2), which is positive infinity, so the right derivative of function at x = 1 does not exist
When substituting the derivative expression, it must be noted that the expression of F (x) must be the section with > 1, because the definition of right derivative is the limit of slope when x tends to 1 from the right of 1, but f (1) should be substituted into the section < 1, because the value of F (1) is defined in this section
The right derivative calculated above is actually the limit of the slope of the straight line connecting (x, f (x)) and (1,1), but this is not the definition of derivative. Because 1 is not equal to f (1), you can see that f '(x) is not right continuous at x = 1

How to draw the derivative image of a function image? How to draw the original function image when given the derivative image? As shown in the figure, given a function image, how to draw its derivative image? As shown in the figure, given a derivative image, how to draw his original function image?

If x > 0, it should be y = - 1 / 2 * x ^ 2 + 4x-3, which is the leftmost part. Because the picture is not clear, it seems that there is an arrow. If there is no y = 3, then X

Derivative function If f '(LNX) = x, then f (x)= ∫ f ′ (x? 2) DX = x ^ 4 + C then f (x)= I want to see the problem solving process

F '(LNX) = x f' (T) = e ^ t two side integral
f(t)=e^t+C
That is, f (x) = e ^ x + C
∫[f′(x^2)]dx = x^4 + C
x^4=∫4x^3dx
4x^3=f′(x^2)
Let x ^ 2 = t
4t^(3/2)=f'(t)
Two side integral f (T) = (8 / 5) T ^ (5 / 2) + C