Let y= x² (x + 1)³ (x + 2)⁴ 
Taking logarithm on both sides we get,
log y = log [x² (x + 1)³ (x + 2)⁴]
= log x² + log (x + 1)³ + log (x + 2)⁴
= 2 log x + 3 log (x + 1) + 4 log (x + 2)
Differentiating with respect to 'x' we have
\(\frac { 1 }{ y } .\frac { dy }{ dx } =\frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \)
\(\frac { dy }{ dx } =y\left[ \frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \right] \)
\(\frac { dy }{ dx } ={ x }^{ 2 }{ \left( x+1 \right) }^{ 3 }{ \left( x+2 \right) }^{ 4 }\left[ \frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \right] \)

Let y = sin (tan^-1 (e^-x))
Differentiating with respect to 'x' we have,
\(\frac { dy }{ dx } =\frac { d }{ dx } \left[ \sin { \left( \tan ^{ -1 }{ \left( { e }^{ -x } \right) } \right) } \right] \)
\(=\cos { \left( \tan ^{ -1 }{ \left( { e }^{ -x } \right) } \right) } .\frac { d }{ dx } \tan ^{ -1 }{ \left( { e }^{ -x } \right) } =\cos { \left( \tan ^{ -1 }{ \left( { e }^{ -x } \right) } \right) } .\frac { 1 }{ 1+{ e }^{ -2x } } .\frac { d }{ dx } \left( { e }^{ -x } \right) \)         
\(=\left[ \because \frac { d }{ dx } ({ tan }^{ -1 }x)=\frac { 1 }{ 1+{ x }^{ 2 } } \right] \)
\(=\cos { \left( \tan ^{ -1 }{ \left( { e }^{ -x } \right) } \right) } .\frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right) \left( -1 \right) =\frac { { -e }^{ -x }.\cos { \tan ^{ -1 }{ \left( { e }^{ -x } \right) } } }{ 1+{ e }^{ -2x } } \)

Given xy + y² = tan x + y
Differentiating with respect to 'x' we get,
\(x.\frac { dy }{ dx } +y\left( 1 \right) +2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow x.\frac { dy }{ dx } +y+2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } \left( x+2y-1 \right) =\sec ^{ 2 }{ x-y } \Rightarrow \frac { dy }{ dx } =\frac { \sec ^{ 2 }{ x-y } }{ x+2y-1 } \)

\(\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 7 }-{ 2x }^{ 5 }+1 }{ { x }^{ 3 }-{ 3x }^{ 2 }+2 } } =\lim _{ x\rightarrow 7 }{ \frac { { x }^{ 7 }-{ x }^{ 5 }-{ x }^{ 5 }+1 }{ { x }^{ 3 }-{ x }^{ 2 }-{ 2x }^{ 2 }+2 } } \)
\(=\lim _{ x\rightarrow 7 }{ \frac { { x }^{ 5 }({ x }^{ 2 }-1)-1({ x }^{ 5 }-1) }{ { x }^{ 2 }(x-1)-2({ x }^{ 2 }-1) } } \)
Dividing the numerator and denominator by (x-1),
\(\\ =\lim _{ x\rightarrow 1 }{ \frac { \frac { { x }^{ 5 }({ x }^{ 2 }-1) }{ x-1 } -1\frac { { (x }^{ 5 }-1) }{ x-1 } }{ \frac { { x }^{ 2 }(x-1) }{ x-1) } -2\frac { ({ x }^{ 2 }-1) }{ x-1 } } } \)
\(=\frac { \lim _{ x\rightarrow 1 }{ { x }^{ 5 }(x+1)-\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 5 }-{ 1 }^{ 5 } }{ x-1 } } } }{ \lim _{ x\rightarrow 1 }{ { x }^{ 2 } } -\lim _{ x\rightarrow 1 }{ 2(x+1) } } \)
\(=\frac { 1(1+1)-5(1)^{ 4 } }{ { 1 }^{ 2 }-2(1+1) } \left[ \because \lim _{ n\rightarrow a }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n.{ a }^{ n-1 } } \right] \)
\(=\frac { 2-5 }{ 1-2(2) } =\frac { -3 }{ -3 } =1\)

Given \(f(x)=\left\{ \begin{matrix} 6ax+3b & if\quad x>1 \\ ax-2b & if\quad x<1 \\ 15 & if\quad x=1 \end{matrix} \right\} \)
\(L[{ f(x)] }_{ x=1 }=\lim _{ x\rightarrow { 1 }^{ - } }{ f(x) =\lim _{ h\rightarrow 0 }{ f(1-h)=\lim _{ h\rightarrow 0 }{ a(1-h)-2b\quad [\because f(x)=ax-2b\quad if\quad x<1] } } } \)
\(=\lim _{ h\rightarrow 0 }{ a-ah-2b=a-2b } \) .....(1)
R[f(x)]_x=1=\(\lim _{ x\rightarrow { 1 }^{ + } }{ f(x)=\lim _{ h\rightarrow 0 }{ f(1+h) } } \)
\(=\lim _{ x\rightarrow { 1 }^{ + } }{ 6a(1+h)+3b\quad [\because f(x)=6ax+3b\quad if\quad x>1] } \)
= 6(91+0) + 3b = 6a + 3b   ....(2)
Also, f(1) = 15 ....(3)
Since f(X) is continuous at x =1,
L[f(x)]_x=1=R[f(x)]_x=1=f(1)
\(\Rightarrow\) a - 2b = 6a + 3b = 15 [using (1), (2) and (3)]
\(\Rightarrow\) a - 2b = 15 ....(4)
and 6a+3b=15 \(\Rightarrow\) 2a + b =5  ..(5)
(4) \(\times\) 2 \(\rightarrow \) 2a - 4b = 30
       -             -      -
(5)        \(\rightarrow \)  2a + b = 5
____________________
              -5b = 25  \(\Rightarrow\) b = -5
Substituting b = -5 in (4) we get,
a - 2(-5) = 15
\(\Rightarrow\) a + 10 = 15
\(\Rightarrow\) a = 15 - 10 = 5
\(\therefore\)    a = 5 and b = -5.