Consider y=2e^x4 log?x
\(\frac{dy}{dx}\)=\(\frac{d}{dx}\) (2e^x4 log?x)
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=2(log?x \(\frac{d}{dx} (e^{x^4})+e^{x^4}\frac{d}{dx}\) (log?x))
\(\frac{dy}{dx}\)=\(2(log?x.e^{x^4}\frac{d}{dx} {x^4}+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2(log?x.e^{x^4}.4x^3+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2e^{x^4} (4x^3 log?x+\frac{1}{x})\)
&re4;\(\frac{dy}{dx}=\frac{(2e^{x^4} (4x^4 log?x+1))}{x}\)

Given function is f(x) = tan x on [ \(\frac {\pi }{4}, \frac {5\pi }{4}\) ]
F(x) = tan x is not defined at x on [ \(\frac {\pi }{4}, \frac {5\pi }{4}\) ]
So, f(x) is not continuous on [ \(\frac {\pi }{4}, \frac {5\pi }{4}\) ].
Hence, Rolle?s orem is not applicable.

Consider y=\(\sqrt{\frac{x+1}{3x-1}}\)
Applying log to both sides, we get
log?y=log?\(\sqrt{\frac{x+1}{3x-1}}\)
log?y=\(\frac{1}{2} log?\left (\frac{x+1}{3x-1}\right )\)
log?y=\(\frac{1}{2}\) (log?(x+1)-log?(3x-1))
Differentiating with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{d}{dx} (log?(x+1))-\frac{d}{dx} (log?(3x-1))\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{1}{x+1}-\frac{3}{3x-1}\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{3x-1-3x-3}{(x+1)(3x-1)})\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{-4}{(x+1)(3x-1)}\right )\)
\(\frac{dy}{dx}\)=\(\sqrt{\frac{x+1}{3x-1}} \left (\frac{-2}{(x+1)(3x-1)}\right )\)
\(\frac{dy}{dx}\)=\(\frac{-2}{(3x-1) \sqrt{(3x-1)(x+1)}}\)

Given that, y=3x²+log?(4x)
\(\frac{dy}{dx}=6x+\frac{1}{4x}.4=6x+\frac{1}{x}=\frac{6x^2+1}{x}\)
\(\frac{d^2 y}{dx^2}=\frac{\frac{d}{dx} (6x^2+1).(x)-\frac{d}{dx} (x).(6x^2+1)}{x^2} \Big(using\, \frac{d}{dx} (\frac{u}{v})=\frac{(\frac{d}{dx} (u).v-\frac{d}{dx} (v).u)}{v^2}\Big)\)
\(\frac{d^2 y}{dx^2}=\frac{(12x.x-6x^2-1)}{x^2} \)
\(\frac{d^2 y}{dx^2}=\frac{6x^2-1}{x^2} = 6-\frac{1}{x^2}\).

Given that, x=6 sin^-1?2t and y=\(\frac{1}{\sqrt{1-4t^2}}\)
\(\frac{dx}{dt}\)=\(\frac{6}{\sqrt{1-4t^2}}.2=\frac{12}{\sqrt{1-4t^2}}\)
\(\frac{dy}{dt}\)=-\(\frac{1}{2(1-4t^2)^{3/2}}.(-8t)=\frac{4t}{(1-4t^2)^{3/2}}\)
\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t}{(1-4t^2)^{3/2}}.\frac{\sqrt{1-4t^2}}{12}\)
\(\frac{dy}{dx}\)=\(\frac{t}{3(1-4t^2)}\)

We differentiate given function with help of chain rule and hence outer function becomes ?sin and inner function is differentiated into 2x, refore answer comes out to be -sin (x²+5).2x.

Given that, x=log?t² and y=\(\frac{1}{t}\)
\(\frac{dx}{dt}\)=\(\frac{1}{t^2}.2t=\frac{2}{t}\)
\(\frac{dy}{dt}\)=-\(\frac{1}{t^2}\)
&re4;\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=-\frac{1}{t^2}.\frac{t}{2}=-\frac{1}{2t}\)

We differentiate given function with help of chain rule and hence
outer function is differentiated into cos, and inner function comes out to be a and constant b becomes 0, which is multiplied to whole function and answer comes out to be
=> a.cos (ax + b).

According to Rolle?s orem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b)
iii) f( a ) = f( b ) n re exists at least one point c ∈ (a,b) such that f?( c ) = 0

Consider y=(log?(log?(x⁵)))
\(\frac{dy}{dx}=\frac{1}{log?x^5} \frac{d}{dx} (log?x^5)\)
\(\frac{dy}{dx}=\frac{1}{log?x^5}.\frac{1}{x^5}.\frac{d}{dx} (x^5)\)
\(\frac{dy}{dx}=\frac{1}{log?x^5}.\frac{1}{x^5}.5x^4\)
&re4;\(\frac{dy}{dx}=\frac{5}{x log?x^5}\)