Question

How do you use the chain rule to differentiate `log_(13)cscx`?

Answer 1

`(dy)/(dx)=-1/ln13*cotx`

Explanation:

Here ,

`y=log_13cscx`

Using Change of Base Formula:

#color(blue)(log_aX=log_k X/log_ka ,where,k " is the new base"#

So .

`y=log_ecscx/log_e13=1/ln13(lncscx)`

Let ,

`y=1/ln13lnu and u=cscx`

`(dy)/(du)=1/ln13*1/u and (du)/(dx)=-cscxcotx`

Using Chain Rule:

`color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx`

`:.(dy)/(dx)=1/ln13*1/u(-cscxcotx)`

Subst. `u=cscx`

`:.(dy)/(dx)=1/ln13*1/(cscx)(-cscxcotx)`

`:.(dy)/(dx)=-1/ln13*cotx`

Answer 2

Same as above answer, different method.

Explanation:

Alternatively...

Let `y = log_13(cscx)`
`rArr y = -log_13(sinx)`
`rArr sinx = 13^-y`

Using implicit differentiation:
`d/dx(sinx) = d/dx(13^-y)`
`rArr d/dx(sinx) = d/dy(e^(-y*ln13))*dy/dx`
`:. cosx = -ln13*13^-y*dy/dx`

But `13^-y = sinx`
`:. dy/dx = cosx/sinx -: -ln13`
`= -cotx/ln13`

As above.