MHB Solve Derivative Questions with Quotient Rule

[lastochka]

Hello, I am struggling with these two questions. I think here should be used a quotient rule, but I am not sure how to proceed.
a) f(x)=sin$\frac{1}{x}$
b)g(x)=$\frac{1}{sinx}$
Can someone please help. Thanks

 

[I like Serena]

Hello, I am struggling with these two questions. I think here should be used a quotient rule, but I am not sure how to proceed.
a) f(x)=sin$\frac{1}{x}$
b)g(x)=$\frac{1}{sinx}$
Can someone please help. Thanks

Hi lastochka,

That should be a chain rule.
It says that the derivative of $f(g(x))$ is $f'(g(x))\cdot g'(x)$.

Applying that to a) we get:
$$\left(\sin \left(\frac{1}{x}\right)\right)'
= \sin' \left(\frac{1}{x}\right) \cdot \left( \frac{1}{x} \right)'
= \cos \left(\frac{1}{x}\right) \cdot (x^{-1})'
= \cos \left(\frac{1}{x}\right) \cdot (-1 \cdot x^{-2})
= - \frac{\cos \left(\frac{1}{x}\right)}{x^2}
$$

How could we apply that to b)? (Wondering)

 

[SuperSonic4]

Hello, I am struggling with these two questions. I think here should be used a quotient rule, but I am not sure how to proceed.
a) f(x)=sin$\frac{1}{x}$
b)g(x)=$\frac{1}{sinx}$
Can someone please help. Thanks

You can use the quotient rule on b if you like although the chain rule works just as well

Let $$u = 1$$ and $$v = \sin(x)$$ and take it from there

 

[lastochka]

Thank you, I like Serena and SuperSonic4!
Question (b) I did with quotient rule:
g$^{\prime}$=$\frac{0sinx-1*cosx}{{sinx}^{2}}$=$\frac{-cosx}{{sinx}^{2}}$

but how to do it with Chain Rule? This fraction confuses me
Thank you again for helping!

 

[Dethrone]

Using the chain rule:

$$\left(\frac{1}{\sin\left({x}\right)}\right)'=[\left(\sin\left({x}\right)\right)^{-1}]'=-1[\sin\left({x}\right)]^{-2}\cos\left({x}\right)=-\frac{\cos\left({x}\right)}{[\sin\left({x}\right)]^2}$$

 

[MarkFL]

Or one could write:

$$\frac{d}{dx}\left(\frac{1}{\sin(x)}\right)=\frac{d}{dx}\left(\csc(x)\right)=-\csc(x)\cot(x)$$ :D

 

[lastochka]

Thank you, Rido12 and MarkFL!