y = 6 sin x log₁₀ x + e
Let u = sin x⇒ u' = cos x
\(v={ log }_{ 10 }x\Rightarrow { v }^{ ' }=\cfrac { 1 }{ x } { log }_{ 10 }e\) 
Now
y = 6uv + e
y' = 6[uv' + vu'] + 0
(i.e.,) \(\cfrac { dy }{ dx } =6\left\{ sinx\cfrac { 1 }{ x } { log }_{ 10 }e+{ log }_{ 10 }x(cosx) \right\} \) 
= \(6\left[ \cfrac { sinx }{ x } { log }_{ 10 }e+cosx+log_{ 10 }x \right] \) 
= \(6\left[ \cfrac { sinx }{ x } { log }_{ 10 }e+cosx+{ log }_{ 10 }e.logx \right] \) 

= \(6{ log }_{ 10 }e\left( \cfrac { sinx }{ x } +cosx.log_{ e }x \right) \)

\(Let\quad u=\log { (1+{ x }^{ 2 } } )\quad and\quad v=\tan ^{ -1 }{ x } \)
\( \Rightarrow \frac { du }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } .\frac { d }{ dx } (1+{ x }^{ 2 })\quad \frac { dv }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\frac { du }{ dx } =\frac { 2x }{ 1+{ x }^{ 2 } } \)

\(\therefore \frac { du }{ dv } =2x\)

\({ x }^{ 2 }+2xy+{ y }^{ 3 }=42\)
Differentiating both sides with respect to 'x' we get,
\(2x+2\left[ x.\frac { dy }{ dx } +y(1) \right] +3{ y }^{ 2 }\frac { dy }{ dx } =0 \Rightarrow 2x+2x\frac { dy }{ dx } +2y+3{ y }^{ 2 }\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } (2x+3{ y }^{ 2 })=-2x-2y \Rightarrow \frac { dy }{ dx } =\frac { -2\left( x+y \right) }{ 2x+3{ y }^{ 2 } } \)

Given xy = 4
Differentiating both sides with respect to 'x' we get,
\(x.\frac { dy }{ dx } =y(1)=0\quad \Rightarrow x\frac { dy }{ dx } =-y ...(1)\)
\(LHS= x\left( \frac { dy }{ dx } +{ y }^{ 2 } \right) =x\frac { dy }{ dx } +x{ y }^{ 2 }= -y+(xy)y=-y+4y\quad \left[ \because \quad xy=4 \right] \)
\(=3y=RHS\)
Hence proved

\(f^{ ' }\left( 1^{ - } \right) =\lim _{ x\rightarrow 1^{ - } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ (x+1) } =1+1=2\quad ...(1)\)
\(f^{ ' }\left( 1^{ + } \right) =\lim _{ x\rightarrow 1^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ x+1 } =1+1=2 ...(2)\)
From (1)and (2), \(f^{ ' }\left( 1^{ - } \right) =f^{ ' }\left( 1^{ + } \right) \)
\(\therefore f\left( x \right) \)is differentiable at x = 1 and \(f^{ ' }\left( 1 \right) =2\)