Calc 1: find tangent to f(x)=5x^2 at x=10

[calculusplease]

Find the equation of the line tangent to the function at the given point.

f(x)=5x^2 at x=10

So I think I understand that the derivative of this function is f(x)=7x, and the slope is 70? That's about as far as I got I could be completely wrong. Any help would be much appreciated.

 

[JeffM]

Find the equation of the line tangent to the function at the given point.

f(x)=5x^2 at x=10





So I think I understand that the derivative of this function is f(x)=7x, and the slope is 70? That's about as far as I got I could be completely wrong. Any help would be much appreciated. You have the power rule correct but not the constant factor rule.

Constant factor rule: $\displaystyle f(x) = a * g(x), where\ a\ is\ a\ constant \implies f'(x) = a * g'(x).$

You added 5 and 2 instead of doing what?

 

[simamura]

Find the equation of the line tangent to the function at the given point.

f(x)=5x^2 at x=10

So I think I understand that the derivative of this function is f(x)=7x, and the slope is 70? That's about as far as I got I could be completely wrong. Any help would be much appreciated.

You applied constant rule incorrectly. f'(x)=5*2*x=10x.

 

[fcabanski]

The derivative of ax^b = b*ax^(b-1)

To find the equation of the tangent line, you need a point. The problem gives you the x coordinate of the point. When you have an x coordinate and an equation, how do you find the y coordinate?

The slope of the tangent line at a point (x,y) is f'(x), which is the derivative of the function at that point.

Once you have the point (x,y) and the slope, you can plug the values into the general line equation (y - y1 = m(x-x1)), then manipulate that to find the slope intercept form of the tangent line equation (y = mx + b).

Here's an example using a different equation.

Find the equation of the line tangent to the function at the given point.

f(x) = 3x^4 at x = 5. (Technically, it's not 'at the given point'. It's at the given x value.)

f(5) = 3*(5^4) = 3*625 = 1875. So the point is (5,1875)

f'(x) = 12x^3, so f'(5) = 1500.

The slope is 1500 at the point (5,1875)

y-y1 = m(x-x1)

y - 1875 = 1500(x - 5)

y -1875 = 1500x - 7500

y = 1500x - 5675