Question

Can someone verify that I took the derivative of this function right.

Answer 1

This the original function

\[y=\sin(\tan \sqrt{1+x^2})\]

Answer 2

this is the derivative I found, it doesn't have to be completely solved

\[\cos(\tan \sqrt{1+x^2})+\sin(\sec \sqrt{1+x^2})(\frac{ 1 }{ 2 }(1+x^2)^{-1/2} (2x))\]

Answer 3

Wrong

Answer 4

You are adding the chains, when you are suppose to multiply them.
Also, you should know that \(\color{#000000}{ \displaystyle \frac{d}{dz} \tan(z)=\sec^2(z) }\)

Answer 5

\(\color{#000000}{ \displaystyle \small \frac{d}{dx} \sin\left[\tan\left(\sqrt{1+x^2}\right)\right] =\cos\left[\tan\sqrt{1+x^2} \right]\times \sec^2\left(\sqrt{1+x^2}\right)\times \frac{1}{2}\left(1+x^2\right)^{-1/2}(2x) }\)

Answer 6

You see what I did?

Answer 7

Note for the tangent function:

\(\color{#000000}{ \displaystyle \small \frac{d}{dx}\tan x=\frac{d}{dx} \left[\frac{\sin x }{\cos x}\right]=\frac{(\sin x)'\cos x-(\cos x)'(\sin x)}{(\cos x)^2} }\)

\(\color{#000000}{ \displaystyle \small =\frac{(\cos x)\cos x-(-\sin x)(\sin x)}{(\cos x)^2}=\frac{\cos^2x+\sin^2x}{\cos^2x} =\frac{1}{\cos^2x} =\sec^2x }\)

Answer 8

oh yes I see, you used the chain rule on the entire function and then you used the chain rule 2 more times in order to solve the firs the first chain rule.

what I tried to do was use the product rule and then use the chain rule 2 times.

Answer 9

Oh, there is product rule in here, really.
You have a function inside a funnction, inside a function ....
so the only way to do this is a chain rule, till you get down to the most inner argument.

Answer 10

if you have any questions, ask.

Answer 11

I have no questions, thanks for helping me out.

Answer 12

Enjoy! GOod luck with differentiation!

Answer 13

Also, in case you didn't know, if you are to differentiate
an absolute value function in a form \(y=\left|f(x)\right|\), then:
\(\color{blue}{\text{------------------------------------------------------------------}}\)

The definition of an absolute value of \(z\) (as long as \(z\) is
a real number,) is as follows:
\(z :=\sqrt{z^2}\)
(The \(:=\) sign reads as, \(«\)by definition equal to\(»\)).

Therefore (let \(y\) be a function of \(x\)), and it by definition
follows that,
\(\color{#000000}{ \displaystyle \frac{d}{dx} \left|y\right|= \frac{d}{dx} \sqrt{y^2 }=\frac{d}{dx} \left(y^2\right)^{1/2} }\)

and this can be computed via a chain rule (twice).
\(\color{#000000}{ \displaystyle =\frac{1}{2}\left(y^2\right)^{-1/2} \times 2y \times y' \ }\)

this can be simplified as follows:
\(\color{#000000}{ \displaystyle =\frac{2y}{2\sqrt{y^2 }} \times y' }\)

Recall the definition of absolute value to simplify
this once more, and cancel the 2's. You will get:
\(\color{#000000}{ \displaystyle =\frac{y}{|y|} \times y' }\)

Answer 14

Therefore you may conclude that
\(\color{#000000}{ \displaystyle \frac{d}{dx}|f(x)|=\frac{f(x)}{|f(x)|} \times f'(x) }\)