Given f(x) = e^x, a = 0, b = 1
f?( c ) = \(\frac {f( b )-f( a )}{b-a}\)
e^c = \(\frac {e-1}{1-0}\)
e^c = e ? 1
C = log\(_e^{e-1}\)

Consider y=4x^ex
Applying log on both sides, we get
log?y=log?4x^ex
log?y=log?4+log?x^ex (?log?ab=log?a+log?b)
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}=0+\frac{d}{dx}(e^x \,log?x)(?log?a^b=b \,log?a)\)
\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \,(e^x) \,log?x+e^{x} \,\frac{d}{dx} \,(log?x)\)
\(\frac{dy}{dx}=y(e^x log?x+\frac{e^x}{x})\)
\(\frac{dy}{dx}=\frac{4x^{e^{x}}e^x \,(x log?x+1)}{x}=4x^{e^x-1} \,e^x \,(x log?x+1)\).

A function is said to be continuous when it is both left continuous and right continuous. Mamatical expression for a function f is right continuous on (a,b) is lim_x?a+?f(x)=f( a ).

Given that, \(y=4x^4+2x\)
\(\frac{dy}{dx}\)=16x³+2
\(\frac{d^2 y}{dx^2}\)=48x²
\(\frac{d^2 y}{dx^2}\)-6 \(\frac{dy}{dx}=48x^2-96x^3-12\)
=12(4x²-8x-1)

We know that derivative of tanx is sec²(x), now in above question we get tan(x+4), hence its derivative comes out to be sec²(x+4), as inside expression (x+4) is differentiated into 1.
Therefore answer is sec²(x+4).

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

Consider y=log?(e^5x3)
y=5x³ (&re4;log?e^x=x)
?\(\frac{dy}{dx}\)=\(\frac{d}{dx} (5x^3)\)
&re4;\(\frac{dy}{dx}=5(3x^2)=15x^2\)

Given that, x=9t⁴ and y=t
\(\frac{dx}{dt}\)=36t³
\(\frac{dy}{dt}\)=1
&re4;\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{36t^3}\)

Lagrange’s mean value orem is also called mean value orem and Rolle?s orem is just a special case of Lagrange’s mean value orem when f( a ) = f( b ).

According to Rolle?s orem, if f : [a,a+h] → R is a function such that
i) f is continuous on [a,a+h]
ii) f is differentiable on (a,a+h)
iii) f( a ) = f(a+h) n re exists at least one &ta; c ∈ (0,1) such that f?(a+&ta;h) = 0