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jech3947
31.01.2020 •
Mathematics
Can someone explain the chain rule?
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Ответ:
\begin{displaymath}\sin(2x) = 2\sin(x) \cos(x) \cdot\end{displaymath}
So we will use the product formula to get \begin{displaymath}\Big(\sin(2x)\Big)' = 2 \Big(\sin'(x) \cos(x) + \sin(x) \cos'(x)\Big)\end{displaymath}
which implies \begin{displaymath}\Big(\sin(2x)\Big)' = 2 \Big(\cos^2(x) - \sin^2(x)\Big)\cdot\end{displaymath}
Using the trigonometric formula $\cos(2x) = \cos^2(x) -\sin^2(x)$, we get \begin{displaymath}\Big(\sin(2x)\Big)' = 2 \cos(2x)\cdot\end{displaymath}
Once this is done, you may ask about the derivative of $\sin(5x)$? The answer can be found using similar trigonometric identities, but the calculations are not as easy as before. Again we will see how the Chain Rule formula will answer this question in an elegant way.In both examples, the function f(x) may be viewed as:
\begin{displaymath}f(x) = h\Big(g(x)\Big)\end{displaymath}
where g(x) = 1+x2 and h(x) = x10 in the first example, and $h(x) = \sin(x)$ and g(x) = 2x in the second. We say that f(x) is the composition of the functions g(x) and h(x) and write \begin{displaymath}f(x) = h \circ g(x).\end{displaymath}
The derivative of the composition is given by the formula \begin{displaymath}\Big(f \circ g(x)\Big)' = g'(x) f'(g(x)).\end{displaymath}
Another way to write this formula is \begin{displaymath}\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\end{displaymath}
where $y = f \circ g(x) = f(u)$ and u = g(x). This second formulation (due to Leibniz) is easier to remember and is the formulation used almost exclusively by physicists.Example. Let us find the derivative of
\begin{displaymath}f(x) = (1+x^2)^{100}\cdot\end{displaymath}
We have $f(x) = h\circ g(x)$, where g(x) = 1+x2 and h(x) = x100. Then the Chain rule implies that f'(x) exists, which we knew since it is a polynomial function, and \begin{displaymath}f'(x) = 2x \cdot \Big[100(1+x^2)^{99}\Big] = 200 x (1+x^2)^{99}\;.\end{displaymath}
Example. Let us find the derivative of
\begin{displaymath}f(x) = \sin(5x)\cdot\end{displaymath}
We have $f(x) = h\circ g(x)$, where g(x) = 5x and $h(x) = \sin(x)$. Then the Chain rule implies that f'(x) exists and \begin{displaymath}f'(x) = 5\cdot \Big[\cos(5x)\Big]= 5 \cos(5x)\cdot\end{displaymath}
In fact, this is a particular case of the following formula
\begin{displaymath}\Big[ f(ax+b)\Big]' = a f'(ax+b) \cdot\end{displaymath} By the way this is not vandalism since i changed it up a little message me if u need link
Ответ:
question 1 is angle addition postulate
Question 2 is 2x=56
Step-by-step explanation: