Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4

Question 1.
(frac{e^{x}}{sin x})
Solution:
Let y = (frac{e^{x}}{sin x}).

where x ≠ nπ, x ∈ Z.

Question 2.
(e^{sin ^{-1} x})
Solution:
Let y = (e^{sin ^{-1} x})
Put sin^-1x = t.
∴ y = e^t.

Question 3.
(e^{x^{3}})
Solution:
Let y = (e^{x^{3}}). Put x³ = t.
∴ y = e^t and t = x³.
⇒ (frac { dy }{ dt }) = e^t and (frac { dt }{ dx }) = 3x².
∴ (frac { dy }{ dx }) = (frac { dy }{ dt }) x (frac { dt }{ dx }) = e^t x 3x² = 3x²(e^{x^{3}}).

Question 4.
sin(tan^-1 e^-x)
Solution:
Let y = sin(tan^-1 e^-x)
Put y = sin s, s = tan^-1 t and t = e^-x

Question 5.
log (cos e^x)
Solution:
Let y = log (cos e^x).
Put y = log s, s = cos t and t = e^x.

Question 6.
e^x + (e^{x^{2}}) + … + (e^{x^{5}})
Solution:

Question 7.
(sqrt{e^{sqrt{x}}}), x > 0
Solution:
Let y = (sqrt{e^{sqrt{x}}}).
Put y = (sqrt{s}), s = e^t and t = (sqrt{x}).
Differentiating, we get

Question 8.
log log x, x > 1
Solution:
Let y = log(log x).
Put y = log t and t = log x.
Differentiating, we get
(frac { dy }{ dt }) = (frac { 1 }{ t }) and (frac { dt }{ dx }) = (frac { 1 }{ x }).
∴ (frac { dy }{ dx }) = (frac { dy }{ dt }) x (frac { dt }{ dx }) = (frac { 1 }{ t }) x (frac { 1 }{ x }) = (frac { 1 }{ log x }) . (frac { 1 }{ x })
= (frac { 1 }{ x log x }), x > 0

Question 9.
(frac { cos x }{ log x }), x > 0
Solution:
Let y = (frac { cos x }{ log x }).

Question 10.
cos (log x + e^x), x > 0
Solution:
Let y = cos (log x + e^x)
Put y = cos t, t = log x + e^x
Differentiating, we get
(frac { dy }{ dx }) = – sin t and (frac { dt }{ dx }) = (frac { 1 }{ x }) + e^x.
∴ (frac { dy }{ dx }) = (frac { dy }{ dt }) x (frac { dt }{ dx }) = – sin t x ( (frac { 1 }{ x }) x e^x).
= – sin (log x + ^x) x ((frac { 1 }{ x }) + e^x)
= – (frac { 1 }{ x }) (1 + xe^x) sin (log x + e^x).