Can you please help me with finding the derivative?

Question

ok, so if you have the equation 3x^2-5x, then what is the equation of the tangent line at x=1

Thanks in Advance.

Answer 1

derivative: 6x-5

x=1 m=6*1-5=1

y=mx+q y=x+q

x=1 Y=3*1^2-5*1=-2

-2=1+q q=-3

tangent y = x -3

Answer 2

First find the derivative of the curve 3x^2-5x => 6x-5 at the point x=1 => 6-5 = 1. This is the slope(m) of the tangent line.
Now, y = mx + b gives us a tangent line
We have y = 3x^2-5x to find y plug x=1 => y = -2
So, y = mx + b becomes -2 = 1x + b => -2 = 1(1) + b => b = -3
Therefore, the tangent line is given by y = x -3.

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Answer 3

y = 3x^2-5x
dy/dx = 6x-5 = slope = 1 at x=1.
When x = 1, y = -2
So equation of tangent line at x =1 is y = mx+b
y = 1x +b
-2 = 1+ b
b = -3
So equation is y = x-3

Answer 4

Derivative rules:

f(x) = g(x) + h(x)

f'(x) = g'(x) + h'(x)

For a function f(x) = ax^n, the derivative f'(x) = anx^(n-1)

The derivative f'(x) is the function of the slope of f(x) at x. So, to find the tangent (the slope) at a point x for your function, derive it, then set x to 1 and solve. That is your tangent (slope).

f(x) = 3x^2 - 5x
f'(x)=3*2*x^1 - 5*1*x^0
= 6x - 5

f'(1) = 6(1) - 5 = 1.

Answer 5

3x^2 - 5x isn't an equation, because it is not set equal to anything, therefore it cannot be solved. However, if it were:

y = 3x^2 - 5x

then you could take the derivative (dy/dx) as:

3(2)x^(2-1) - 5(1)x^(1-1)
6x - 5

Hope this helps!

Answer 6

f(x) = 3x² - 5x
f `(x) = 6x - 5
f `(1) = 1
f (1) = - 2
y + 2 = 1.(x - 1)
y = x - 3 is equation of required tangent line.

Answer 7

Answer: Tangent line is y=x-3.

f(x)=3*x^2-5*x= -2 for x=1
f'(x)=6*x-5=m=1 for x=1
The point (1, -2) satisfies f(x), f'(x), tangent line.