# GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4

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 = et.

Question 3.
$$e^{x^{3}}$$
Solution:
Let y = $$e^{x^{3}}$$. Put x³ = t.
∴ y = et and t = x³.
⇒ $$\frac { dy }{ dt }$$ = et and $$\frac { dt }{ dx }$$ = 3x².
∴ $$\frac { dy }{ dx }$$ = $$\frac { dy }{ dt }$$ x $$\frac { dt }{ dx }$$ = et 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 ex)
Solution:
Let y = log (cos ex).
Put y = log s, s = cos t and t = ex.

Question 6.
ex + $$e^{x^{2}}$$ + … + $$e^{x^{5}}$$
Solution:

Question 7.
$$\sqrt{e^{\sqrt{x}}}$$, x > 0
Solution:
Let y = $$\sqrt{e^{\sqrt{x}}}$$.
Put y = $$\sqrt{s}$$, s = et 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 + ex), x > 0
Solution:
Let y = cos (log x + ex)
Put y = cos t, t = log x + ex
Differentiating, we get
$$\frac { dy }{ dx }$$ = – sin t and $$\frac { dt }{ dx }$$ = $$\frac { 1 }{ x }$$ + ex.
∴ $$\frac { dy }{ dx }$$ = $$\frac { dy }{ dt }$$ x $$\frac { dt }{ dx }$$ = – sin t x ( $$\frac { 1 }{ x }$$ x ex).
= – sin (log x + x) x ($$\frac { 1 }{ x }$$ + ex)
= – $$\frac { 1 }{ x }$$ (1 + xex) sin (log x + ex).