Procrastination.
I should stop doing that. It's really not good for my mental health. Anyways, this week we discussed, surprise, surprise, derivatives! Apparently there even more properties of derivatives to be learned! (That was not sarcasm, I promise you.)
Our focus this week was on the derivatives of composite functions and the chain rule. As we discovered through part of our activity on Monday, the derivative of a composite function at a point equals the composition of the equations of the tangent lines. The chain rule is used to the derivative of such functions; it states: the derivative of a composite function f(g(x)) equals f'(g(x))*g'(x). Essentially, the derivative is the derivative of the outside function with the inside unchanged multiplied by the derivative of the inside function. This concept isn't particularly hard to understand, but I had a somewhat difficult time learning it because I was absent the day of this lecture and activity. That tends to be a hindrance on one's learning. However, thanks to Cresswell's completely legible and not at all incoherent notes, I was able to catch up quickly. In all seriousness, though, I had only a few problems understanding the chain rule but as soon as I got the notes, I got it.
Also, there are instances where two or more chain rules have to be done, like in the case of f(x)=(cos(3x^2))^2, which also made life a bit harder. The first step to solving this problem, or at least, the first step that helps me, is to identify the different functions that makes up the composition. For f(x), it would be x^2 (the outermost layer), cos(x) (the middle layer), and 3x^2 (the innermost layer). Once all parts have been identified, just follow the chain rule!
Take the derivative of the outermost layer, leaving the inside intact:
f'(x)=2(cos(3x^2))
Then multiply that by the derivative of the next innermost layer:
f'(x)=2(cos(3x^2))*(-sin(3x^2))
Finally, multiply that by the derivative of the ultimate innermost layer:
f'(x)=2(cos(3x^2))*(-sin(3x^2))*(6x)
And that's your derivative!
It seems like a long and complicated process, but once the basics are down, it is relatively easy. I just make it hard on myself by *sigh* procrastinating.
I should stop doing that. It's really not good for my mental health. Anyways, this week we discussed, surprise, surprise, derivatives! Apparently there even more properties of derivatives to be learned! (That was not sarcasm, I promise you.)
Our focus this week was on the derivatives of composite functions and the chain rule. As we discovered through part of our activity on Monday, the derivative of a composite function at a point equals the composition of the equations of the tangent lines. The chain rule is used to the derivative of such functions; it states: the derivative of a composite function f(g(x)) equals f'(g(x))*g'(x). Essentially, the derivative is the derivative of the outside function with the inside unchanged multiplied by the derivative of the inside function. This concept isn't particularly hard to understand, but I had a somewhat difficult time learning it because I was absent the day of this lecture and activity. That tends to be a hindrance on one's learning. However, thanks to Cresswell's completely legible and not at all incoherent notes, I was able to catch up quickly. In all seriousness, though, I had only a few problems understanding the chain rule but as soon as I got the notes, I got it.
Also, there are instances where two or more chain rules have to be done, like in the case of f(x)=(cos(3x^2))^2, which also made life a bit harder. The first step to solving this problem, or at least, the first step that helps me, is to identify the different functions that makes up the composition. For f(x), it would be x^2 (the outermost layer), cos(x) (the middle layer), and 3x^2 (the innermost layer). Once all parts have been identified, just follow the chain rule!
Take the derivative of the outermost layer, leaving the inside intact:
f'(x)=2(cos(3x^2))
Then multiply that by the derivative of the next innermost layer:
f'(x)=2(cos(3x^2))*(-sin(3x^2))
Finally, multiply that by the derivative of the ultimate innermost layer:
f'(x)=2(cos(3x^2))*(-sin(3x^2))*(6x)
And that's your derivative!
It seems like a long and complicated process, but once the basics are down, it is relatively easy. I just make it hard on myself by *sigh* procrastinating.