AP®︎/College Calculus AB
- Proof: Differentiability implies continuity
- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof
Using the fact that the derivative of sin(x) is cos(x), we use visual aides to show that the derivative of cos(x) is -sin(x).
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- At1:52, the blue and red curves on the upper graph were shifted by pi/2 to the left, i.e. to the negative side. Why would the formulae on the bottom graph be +pi/2 instead of -pi/2? Thanks!(1 vote)
- Think of the function (x-2)², which is x² shifted left or right by 2.
Look at the vertex of x²; when x=0, then x²=0. So if we want to locate the vertex of (x-2)², we need to make the input of x² (which is now x-2 instead of x) be 0.
So we set x-2=0, or x=2. So the vertex now sits at x=2, so the parabola has been shifted to the right.
In general, replacing x by x-a shifts right by a units, and replacing x by x+a shifts left by a units, because we understand this by setting the input equal to 0 and solving, which flips the sign.(1 vote)
- Is there an algebraic proof for the derivative of cos(x)?(1 vote)
- [Instructor] What I wanna do in this video is make a visual argument as to why the derivative with respect to x of cosine of x is equal to negative sine of x. And we're gonna base this argument, based on a previous proof we made that the derivative with respect to x of sine of x is equal to cosine of x. So we're gonna assume this over here. I encourage you to watch that video. That's actually a fairly involved proof that proves this, but if we assume this, I'm gonna make a visual argument that this right over here is the derivative with respect to x of cosine of x is negative sine of x. So right over here we seen sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph. And we've got an intuition for that in previous videos. Now what I'm gonna do next, is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two and I'm also gonna shift the blue graph to the left by pi over two. And so what am I going to get? Well the blue graph is gonna look like this one right over here and if it was cosine of x up here, we can now say that this is equal to y is equal to cosine of x plus pi over two. This is the blue graph, cosine of x, shifted to the left by pi over two. And this is y is equal to sine of x plus pi over two. Now the visual argument is, all I did, is I shifted both of these graphs to the left by pi over two. So it should still be the case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with respect to x of the red graph, sin of x plus pi over two that that is equal to the blue graph. That that is equal to cosine of x plus pi over two. Now what is sin of x plus pi over two? Well that's the same thing as cosine of x. You can see this red graph is the same thing as cosine of x. We know that from our trig identities and you can also see in intuitively or graphically just by looking at these graphs. Now what is cosine of x plus pi over two? Well once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument. Just start with this knowledge, shift both of these graphs to the left by pi over two, it should still be true, that the derivative with respect to x of sine of x plus pi over two is equal to cosine of x plus pi over two. And this is the same thing as saying what we have right over here. So now we should feel pretty good. We proved this in a previous video and we have a very strong visual argument for this for cosine of x in this video.