I am a collect freshman taking Calculus I. Having problems with this.
Find dy/dx given x^3sin^3(X^3). Please help. I do not know where to start.
3x^2 * ??? Sin Cos ???
A collect freshman? Does that mean I have to pay your tuition if I answer?
The mechanics of finding a derivative of a complicated function frequently involve, implicitly or explicitly, the
CHAIN RULE.
Here is the basic idea, and you use it over and over and over again. You should have memorized the derivatives of about 20 standard functions. You take the complicated function and break it down into simpler functions until every function is one of the standard ones. Then you start taking derivatives of the simpler functions. That's it.
Let's take a fairly simple example:
\(\displaystyle y = x^3sin(x).\) That is not a standard function, but it is a product of two functions, and products of functions ARE standard.
So \(\displaystyle u = x^3\ and\ v = sin(x) \implies y = x^3sin(x) = uv \implies \dfrac{dy}{dx} = u * \dfrac{dv}{dx} + \dfrac{du}{dx} * v.\)
And \(\displaystyle u = x^3 \implies \dfrac{du}{dx} = 3x^2\ and\ v = sin(x) \implies \dfrac{dv}{dx} = cos(x).\)
Therefore \(\displaystyle y = x^3sin(x) = uv \implies\)
\(\displaystyle \dfrac{dy}{dx} = u * \dfrac{dv}{dx} + \dfrac{du}{dx} * v = x^3 * cos(x) + (3x^2) * sin(x) = x^2\{xcos(x) + 3sin(x)\}.\)
We took an ugly function and broke it down into simpler functions whose derivatives were already known. That example used the product rule, power rule, and sine rule.
Next example is a bit more complex.
\(\displaystyle y = sin^2(3x + 1).\) Not standard.
\(\displaystyle u = 3x + 1 \implies y = sin^2(u).\) Simpler. but still not standard.
\(\displaystyle v = sin(u) \implies y = sin^2(u) = v^2.\) Now that is standard.
\(\displaystyle y = v^2 \implies \dfrac{dy}{dv} = 2v = 2sin(u) = 2sin(3x + 1).\)
That was simple; we used the power rule, but it did not get us what we want, which is \(\displaystyle \dfrac{dy}{dx}.\) We need the chain rule.
In this case, the chain has three links because we made two substitutions:
\(\displaystyle y = sin^2(3x +1) = sin^2(u) = v^2 \implies \dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx}.\)
\(\displaystyle v = sin(u) \implies \dfrac{dv}{du} = cos(u) = cos(3x + 1)\ and\ u = 3x + 1 \implies \dfrac{du}{dx} = 3.\)
Put it all together and you get:
\(\displaystyle y = sin^2(3x + 1) = sin^2(u) = v^2 \implies \dfrac{dy}{dx} = 2sin(3x + 1) * cos(3x + 1) * 3 = 6sin(3x + 1)cos(3x + 1).\)
With this generic explanation of the process and the hint given in post 2, how far can you go?
