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

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Answer 1 · 2018-08-12T11:05:19

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

Here ,

#y=log_13cscx#

#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#

#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 · 2018-08-14T07:43:27

Same as above answer, different method.

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.