Parents can use S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(j) to provide additional support to their children.
S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(j)
Differentiate :
Question 1.
x² w.r.t. x³
Solution:
Let y = x² and z = x³
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\).
Diff. eqn. (1) & (2) w.r.t. x; we have
Question 2.
x³ – x² – x + 1 w.r.t. 3x² – x + 2
Solution:
Let y = x³ – x² – x + 1 … (1)
and z = 3x² – x + 2 …(2)
So we want to find \(\frac { dy }{ dz }\).
Question 3.
sin³ x w.r.t. cos³x
Solution:
Let y = sin³ x … (1)
and z = cos³x … (2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\).
Diff. eqn. (1) & (2) w.r.t. x; we have
Question 4.
sin x³ w.r.t sec² x²
Solution:
Let y = sin³ x … (1)
and z = sec² x² … (2)
So we want to diff. y w.r.t, x i.e. to find \(\frac { dy }{ dz }\).
Diff. eqn. (1) & (2) w.r.t. x; we have
Question 5.
Differentiate
(i) tan-1\(\frac{2 x}{1-x^2}\) w.r.t tan-1
(ii) tan-1\(\frac{x}{\sqrt{1-x^2}}\) w.r.t sec-1\(\frac{1}{2 x^2-1}\)
(iii) tan-1\(\frac{2 x}{1-x^2}\) w.r.t sin-1\(\frac{2 x}{1+x^2}\)
(iv) tan-1\(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t tan-1 x.
Solution:
(i) Given y = tan-1\(\frac{2 x}{1-x^2}\) … (1)
and z = tan-1x … (2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\)
put x = tan θ ⇒ θ = tan-1x
(ii) Given y = tan-1\(\frac{x}{\sqrt{1-x^2}}\) … (1)
and z = sec-1\(\frac{1}{2 x^2-1}\) … (2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\)
putting x = sin θ in eqn. (1); we get
(iii) Given y = tan-1\(\frac{2 x}{1-x^2}\) … (1)
and z = sec-1\(\frac{2 x}{1+x^2}\) … (2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\)
putting x = tan θ ⇒ θ = tan-1x in eqn. (1); we have
(iv) Given y = tan-1\(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) … (1)
and z = tan-1 x … (2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dz }\)
putting x = tan θ ⇒ θ = tan-1x in eqn. (1); we have
