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For example, here we have $y = \sin(x^2 + 5x -6)$.

So, it’s answer will be $\cos(t)\cdot(2x+5)$, where $t= x^2 + 5x -6$. But I want to know what if we solve it directly I.e $\cos(2x+5)$ wrong? What’s the reason? Why choose the 1st approach?

DMcMor
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Rider
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    It is not true that the derivative of $f\circ g(x)$ is $f\circ g'(x)$. – lulu Apr 09 '21 at 16:25
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    Does this answer your question? how to prove the chain rule? – 5201314 Apr 09 '21 at 16:29
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    Perhaps one question you want to think about is this. If the answer is $\cos(x^2+5x-6) \cdot (2x+5)$, could the answer also be $\cos(2x+5)$? If both are correct then you get an equation $$\cos(x^2+5x-6) \cdot (2x+5) = \cos(2x+5)$$ Could that equation really be true? For example, if you plug in $x=0$ radians you get the equation $$\cos(-6) \cdot 5 = \cos(5)$$ If you want, use a calculator to check whether this equation is true. But remember to set it to radians first. – Lee Mosher Apr 09 '21 at 17:07
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    The reason we need to use the chain rule is that if we use it correctly it gives the correct derivative. Your "direct" differentiations are wrong. – David C. Ullrich Apr 09 '21 at 17:17

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If your "formula" worked, then $\frac{d(\sin(2x))}{dx}$ would be $\cos(2)$. Now $\cos(2)$ is a constant. But the derivative gives the slope of a function. So the slope of $y=\sin(2x)$ would be a constant which does not seem to be right when we look at its graph.

MasB
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Even if you know $D(\sin x) = \cos x$ and that $D(x^2+5x-6) = 2x + 5$ , why should you think that $D(\sin (x^2+5x-6)) = \cos(2x+5)$? You need to establish (not choose) the right rule to deal with the derivative of a composition and it turns out it's $$D(f(g(x)) =D(f(t)_{t=g(x)})D(g(x))$$ and so you use the derivative of the components in another way.

Tortar
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Correct me if I'm wrong, but you seem to be asking for an explanation for why we need the chain rule, not for a proof of why it's true. Here is a very non-technical explanation for why we would expect to need a chain rule.

Let's imagine we have an old western-style train robber running along the top of a train. The robber runs $528 \text{ ft}$ for each mile that the train travels, and the train is moving at $30 \text{ mph}$ with respect to the ground. How fast, in $\text{ft}/\text{hr}$ is the robber moving with respect to the ground?

We can do this without any calculus, just paying attention to the units: $$528 \frac{\text{ft}}{\text{mi}} \times 30\frac{\text{mi}}{\text{hr}} = 15840 \frac{\text{ft}}{\text{hr}}.$$

Now, we could also do this with calculus. Let $F(m) = 528m$ be the feet the robber runs per mile the train moves, and let $M(h) = 30h$ be the number of miles the train travels per hour. Then the total distance traveled by the robber after $h$ hours is given by $$F(M(h)) = 528(30h) = 15840h.$$ We can find the speed of the robber in $\text{ft}/\text{hr}$ by differentiating $F(M(h))$. We can do this directly, using the simplified for $15840h$, or we can use the chain rule.

Direct way: $$\frac{d}{dh}F(M(h)) = \frac{d}{dh}15840h = 15840 \frac{\text{ft}}{\text{hr}}.$$

Chain rule: $$\frac{d}{dh}F(M(h)) = F'(M(h))\cdot M'(h) = 528 \cdot 30 = 15840 \frac{\text{ft}}{\text{hr}}.$$

Note that the chain rule way actually corresponds directly to how we computed this without calculus. In this case every function was linear, so everything is greatly simplified, but in more complicated cases the analogy still (losely) holds, that you need both the rate of change of the outside function and the rate of change of the inside function multiplied together.

DMcMor
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