The tangent equation for the curve y x 3 1 at point 1,1

The curve y=1 x 2 can be converted to the power of minus thirds of y=x.

Derivative: y'=

2 3) x to the minus five-thirds power.

So substituting x=1 into y' gives the tangent equation of y'=-(2 3) so the curve y=1 x 2 at the point (1,1).

The slope of is k=-(2 3).

From the point oblique formula: y-1=-(2 3)(x-1), that is, 2x+3y-5=0

First, find the derivative of the curve to get y'=3x squared, and then bring in the tangent abscissa to get k=3, so the tangent equation is y=3(x

Then it's a matter of simplifying to get y3x

Doing this kind of problem is equivalent to finding the equation of a straight line, and the point on the straight line is known, now you only need to know the slope of the straight line, isn't it? So take this curve as a derivative.

y'=3x^2

Put the thought x=-1 generation into it, and get y'=3

So the slope of this straight line is 3

And the dot (-1,1).

Get. y=3x+4

Let the tangent equation for the curve y=x -x+3 be y1=kx1+b, then.

k=y'=(x³-x+3)'=3x²-1

The slope k at this point (1,3) is.

The intercept b of the tangent equation of k=2 is .

b=3-2×1=1

So the tangent equation at the point (1,3) is.

y1=2x1+1

First; 2b, point a(-1, let the tangent of a be tangent to the curve of b(b, b tangent equation: yb 3x

The tangent slope of the crossing b k3b

y'3b²(xb)

a on the tangent; 3b²

There is a real root, about b

b³3b²(-1b)

b 3b 3b ,1) is not on the curve (x1,y

y=x^3+x+1

y'=3x²+1

The slope of the tangent at the point (1,3) k=3 1 +1=4 curve y=x 3+x+1 The tangent equation at the point (1,3) is: y-3=4(x-1).

That is, y=4x-1

Summary. Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

Tangent equation for the curve y=2-3x x at the point (1,-1).

Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

What's that curve of yours? <>

**Question (2).

y'=ln3*3^x

According to the tangent equation y-y0=f'(x0)(x-x0) gives y-3=ln3*3*(x-1).

So the tangent equation laughs and quarrels with a basis of y=3(ln3)(x-1)+3

Derivative of the curve equation: y'=6x^2

The tangent slope of the curve at points (1,3) is: k=y'|x=1=6

Therefore, the tangent equation is: y-3=6(x-1), i.e., y=6x-3

Solution: First search for the curve y=x 3.

Y derivative = 3x 2

It is found according to the general equation of the straight line y=kx+b.

y=kx+b, through (1,1) cut k=3, b=-2 with people, the curve is arranged in y=x3, and the tangent square celery is in the (1,1) point.

is y=3x-2

y=x (-2 3), derivative=

2 3x (-5 3), the tangent slope of the eggplant k=

2 3, so y-1=

2 3 (x-1), the tangent party knows the process is 2x+3y-5=0

Answer: y=3x-2 itself is a straight line through the point (1,1) there is no tangent line, and its tangent line is the straight imitation of the envy line itself.

y=(3^x)-2

Lead-seeking. y'(x)=(3^x)'-0

3^x)ln3

y'(1)=3ln3

So: at point (1,1).

Tangent equations. is y-1=3ln3(x-1).

∵y=x^3-1

y'=3x²

then y'(1)=3 is the slope of the tangent at the point (1,0) The tangent equation is: y-0=3(x-1), that is, 3x-y-3=0 and the slope of the normal is -1 3

The normal equation is: y-0=-1 3*(x-1), that is, x+3y-1=0 [middle school students' mathematics, physics and chemistry] team will answer for you! I wish you progress in your studies, and you can ask if you don't understand!

Tangent: The derivative is y'=3x 2, at x=1, the derivative is 3So the tangent slope is 3, and with the point slope formula, the tangent equation is y= 3(x-1).

Normal: The slope of the normal is -1 3The same point oblique, y=-1 3 (x-1).