Question

If
$y = {{tan(x)}^{tan(x)}}^{tan(x)}$
Then find dy/dx at x=pi/4 ​

Answer 1

Answer:

y=tanxtanxtanx  

Let  m=tanxtanx  

y=tanxm  

Applying logarithmic differentiation as the given function is of the form of a function raised to the power of another function

logy=m∙log(tanx)  

Differentiating on both sides with respect to  x  

1y∙dydx=m∙log(tanx)  

In the RHS of the equation we need to apply product rule

1y∙dydx=1tanx×sec2x.m  

+log(tanx)×dmdx  

dydx=y(1tanx×sec2x.m+log(tanx)×dmdx)⋯(1)  

ddx(logx)=1x  

ddx(tanx)=sec2x  

Now, dmdx  

m=tanxtanx  

Applying logarithmic differentiation,

logm=logtanxtanx  

logm=tanx.log(tanx)  

Differentiating on both sides with respect to  x  

1m∙dmdx=sec2x∙log(tanx)+1tanx×sec2x×tanx  

(uv)′=u′v+v′u  

dmdx=m[sec2x∙log(tanx)+sec2x]  

dmdx=tanxtanx  ∙sec2x[1+log(tanx)]⋯(2)  

Substituting  (2)  in  (1),  

We obtain,

dydx=tanxtanxtanx(sec2x.tanxtanx−1+tanxtanxsec2x(1+log(tanx))

Step-by-step explanation:

y=tanxtanxtanx

Let u=tanx→y=uuu

From the chain rule, dydx=dydu⋅dudx

dudx is sec2x, however dydu is more complicated.

Consider a=uuu

lna=lnuuu

lna=uulnu

Differentiating implicitly:

1adadu=[uu]′lnu+uu[lnu]′

[uu]′ is also not simple, so consider b=uu

lnb=lnuu

lnb=ulnu

Differentiating implicitly:

1bdbdu=lnu+1

dbdu=b(lnu+1)

dbdu=uu(lnu+1)

∴[uu]′=uu(lnu+1)

Therefore1adadu=uu(lnu+1)lnu+uuu

1adadu=uu(lnu+1)lnu+uu−1

dadu=a(uuln2u+uulnu+uu−1)

dadu=uuu(uuln2u+uulnu+uu−1)

Therefore [uuu]′=uuu(uuln2u+uulnu+uu−1)

Therefore dydx=sec2x(uuu(uuln2u+uulnu+uu−1))

Substituting back in:

dydx=sec2xtanxtanxtanx(tanxtanxln2tanx+tanxtanxlntanx+tanxtanx−1