Byrgg
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I missed the lessons at school for derivatives, and I'm not really understanding things from my textbook very well, so I'm hoping someone here can help me a bit with these problems.
1. In each case, find the derivative dy/dx
a) y = 6 - 7x
b) y = {x + 1}/{x - 1}
c) y = 3x^2
The book didn't go into too much detail on the dy/dx thing, and so I don't really have any idea on what to do.
2. Find an equation of the straight line that is tangent to the graph of f(x) = \sqrt{x + 1} and parallel to x - 6y + 4 = 0.
I figured that I could at least find the slope of the line, since it's parallel to x - 6y + 4 = 0. I calculated the slope from this to be 1/6, and since the lines are parallel, then the slope of the tangent to the graph of f(x) = \sqrt{x + 1}[/itex] must also be 1/6. I didn't really know what to do after this.
3. For each function, use the definition of the derivative to dtermine dy/dx, where a, b, c, and m are constants.
a) y = c
b) y = x
c) y = mx + b
c) y = ax^2 + bx + c
I was thinking that I might have to substitute the y in each case for the y in dy/dx, but I'm not really sure, as I said earlier, the book didn't go into dy/dx all that much.
1. In each case, find the derivative dy/dx
a) y = 6 - 7x
b) y = {x + 1}/{x - 1}
c) y = 3x^2
The book didn't go into too much detail on the dy/dx thing, and so I don't really have any idea on what to do.
2. Find an equation of the straight line that is tangent to the graph of f(x) = \sqrt{x + 1} and parallel to x - 6y + 4 = 0.
I figured that I could at least find the slope of the line, since it's parallel to x - 6y + 4 = 0. I calculated the slope from this to be 1/6, and since the lines are parallel, then the slope of the tangent to the graph of f(x) = \sqrt{x + 1}[/itex] must also be 1/6. I didn't really know what to do after this.
3. For each function, use the definition of the derivative to dtermine dy/dx, where a, b, c, and m are constants.
a) y = c
b) y = x
c) y = mx + b
c) y = ax^2 + bx + c
I was thinking that I might have to substitute the y in each case for the y in dy/dx, but I'm not really sure, as I said earlier, the book didn't go into dy/dx all that much.