Calculus with Trig Functions

We got our tests back today. Whew! I got a 92. It’s my first A without test corrections.. but you know what, I might just be a dill hole and do test corrections anyway, because you know what sounds better than a 92? A 97.

I feel that I have been slacking on my blog. I have found myself lost for what to post, but I know our portfolio is due this Friday, so I am going to do my best not to drop the ball as we near the finish line.

So on to the subject at hand:

Calculus with Trig Functions:

I felt like today we had a lot of confused people in class. Perhaps it’s just the fact that we haven’t worked with “h” in a while, but I was shocked at the bombardment of questions at our Professor today. She was showing us exactly how she got the derivative of the sin, tan, cos, cot, sec and csc. It made sense to me. The reason we don’t need to derive it on the test is because we can now memorize the derivative of each trig function and use it (I imagine) with more difficult forms of each function.

I’m going to walk through how we get the derivative of one or two of these functions, and then I’m going to make a pretty little chart of memorization for myself to refer to for the FINAL.

Find the Derivative:

$f(x)=sinx$

$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$\lim_{h\to0}\dfrac{sin(x+h)-sinx}{h}$

One of the rules our professor provided is really key in this next step:

sin(a+b) = sinacosb + sinbcosa (this will be provided for the exam so we don’t have to memorize this particular rule)

$\lim_{h\to0}\dfrac{sinxcosh+sinhcosx-sinx}{h}$

$\lim_{h\to0}(\dfrac{sinxcosh-sinx}{h}+\dfrac{sinhcosx}{h})$ – here we simply separated the equation out so that we could factor out the sinx.

$\lim_{h\to0}\dfrac{sinx(cosh-1)}{h}+\lim_{h\to0}\dfrac{sinhcosx}{h}$

$sinx\lim_{h\to0}\dfrac{cosh-1}{h}+cosx\lim_{h\to0}\dfrac{sinh}{h}$

We were able to factor out the sinx and cos x here because they did not affect h, which is what we are trying to find the limit of.

Earlier we had already solved for the following equations: $\lim_{x\to0}\dfrac{sinx}{x}=1$ and $\lim_{x\to0}\dfrac{cosx-1}{x}=0$

we can now plug those into the equation we have, and get the following:

$(sinx)(0)+(cosx)(1)$

$\frac{d}{dx}sinx=cosx$

See? It’s just using the earliest method we learned to find the derivative of these trig functions. If you can’t remember that method, please read this blog post of mine.

So here’s a list of the need to memorize derivatives:

$\dfrac{d}{dx}(sinx)=cosx$

$\dfrac{d}{dx}(cosx)=-sinx$

$\dfrac{d}{dx}(tanx)=sec^2x$

$\dfrac{d}{dx}(cotx)=-csc^2x$

$\dfrac{d}{dx}(secx)=secxtanx$

$\dfrac{d}{dx}(cscx)=-cscxcotx$

Here’s a list of useful trig identities:

$sin^2x+cos^2x=1$

$sin^2x$ is shorthand for $(sinx)^2$

$tanx=\dfrac{sinx}{cosx}$

$cotx=\dfrac{cosx}{sinx}$

$secx=\dfrac{1}{cosx}$

$cscx=\dfrac{1}{sinx}$

sin (a+b) = sin(a)cos(b) + sin(b)cos(a)

cos (a+b) = cos(a)cos(b) + sin(a)sin(b)

*also remember to keep your calculator in radians for trig functions
That’s all I’ve got for you folks today. I am (confusingly) excited to see what our last Calculus class entails. Then on to Calc 2 in a month!