How come we say d/dy*d/dx?

**Paze** · February 28th 2013, 07:05 PM

So if we are differentiating the y-side of the function, it goes something like this: Let the function be tan(y)=x $\frac{d}{dy}tany\cdot \frac{d}{dx}=1$

I'm having problems understand why it is done like this and it is not touched upon in my lectures. It just seems to be taken for granted.

Why do we differentiate the y-side and then multiply it by y' ? And what IS y' ?! Shouldn't it be the same thing as d/dy of the y-side?

Hope I'm not too confusing!

Thanks!

**Esteban** · February 28th 2013, 07:11 PM

y' is the derivative of y... and this term appears because you have to derivate using the chain rule. For further information check "implicit derivative"

**Paze** · February 28th 2013, 07:21 PM

Right. We are differentiating a function inside a function (which is relevant to your avatar if I'm not mistaken? lol). Thanks!

**Prove It** · February 28th 2013, 07:36 PM

Originally Posted by Paze

So if we are differentiating the y-side of the function, it goes something like this: Let the function be tan(y)=x $\frac{d}{dy}tany\cdot \frac{d}{dx}=1$

I'm having problems understand why it is done like this and it is not touched upon in my lectures. It just seems to be taken for granted.

Why do we differentiate the y-side and then multiply it by y' ? And what IS y' ?! Shouldn't it be the same thing as d/dy of the y-side?

Hope I'm not too confusing!

Thanks!

You need to understand that $\displaystyle \begin{align*} \frac{d}{dx} \left( \textrm{something} \right) \end{align*}$ or $\displaystyle \begin{align*} \frac{d}{dy} \left( \textrm{something} \right) \end{align*}$ are OPERATORS, which mean "take the derivative with respect to the variable..." while $\displaystyle \begin{align*} \frac{dy}{dx} \end{align*}$ means "the derivative of y has been taken with respect to x". Another way to write it would be $\displaystyle \begin{align*} \frac{d}{dy} \left( x \right) \end{align*}$ .

Now as for your other question, what is happening is FIRST to note that y is a function of x, which means that both sides can be derived with respect to x.

$\displaystyle \begin{align*} \tan{(y)} &= x \\ \frac{d}{dx} \left[ \tan{(y)} \right] &= \frac{d}{dx} \left( x \right) \\ \frac{d}{dx} \left[ \tan{(y)} \right] &= 1 \end{align*}$

Then, because we know that y is a function of x, it has then been composed with the tangent function, and so to differentiate we need to apply the chain rule. The "inner" function is y, so its derivative is $\displaystyle \begin{align*} \frac{dy}{dx} \end{align*}$ , and the "outer" function is $\displaystyle \begin{align*} \tan{(y)} \end{align*}$ , so its derivative is $\displaystyle \begin{align*} \frac{d}{dy} \left[ \tan{(y)} \right] \end{align*}$ . This gives

$\displaystyle \begin{align*} \frac{d}{dx} \left[ \tan{(y)} \right] &= 1 \\ \frac{d}{dy}\left[ \tan{(y)} \right] \frac{dy}{dx} &= 1 \end{align*}$

Hope that helps...

**Paze** · February 28th 2013, 08:32 PM

Yep! That's how I came to understand it. Thank you both both very much.

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