Differentiation Class 12 OP Malhotra Exe-8J ISC Maths Solutions Ch-8 Solutions. In this article you would learn about differentiation of a function with respect to another function . Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.
Differentiation Class 12 OP Malhotra Exe-8J ISC Maths Solutions Ch-8
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-8 | Differentiation |
| Writer | OP Malhotra |
| Exe-8(J) | differentiation of one function with respect to another function |
Exercise- 8J
Differentiation Class 12 OP Malhotra Exe-8J Solution
Que-1: x² w.r.t. x³
Sol: Let y = x² and z = x³
So we want to diff. y w.r.t. z i.e. to find dy/dz.
Diff. eqn. (1) & (2) w.r.t. x; we have
Que-2: x³ – x² – x + 1 w.r.t. 3x² – x + 2
Sol: Let y = x³ – x² – x + 1 … (1)
and z = 3x² – x + 2 …(2)
So we want to find dy/dz.
Que-3: sin³ x w.r.t. cos³x
Sol: Let y = sin³ x … (1)
and z = cos³x … (2)
So we want to diff. y w.r.t. z i.e. to find dy/dz.
Diff. eqn. (1) & (2) w.r.t. x; we have
Que-4: sin x³ w.r.t sec² x²
Sol: Let y = sin³ x … (1)
and z = sec² x² … (2)
So we want to diff. y w.r.t, x i.e. to find dy/dz.
Diff. eqn. (1) & (2) w.r.t. x; we have
Que-5: Differentiate
(i) tan-1(2x/1-x²) w.r.t tan-1
(ii)tan-1(x/√1-x²) w.r.t sec-1(1/2x²-1)
(iii) tan-1(2x/1-x²) w.r.t sin-1(2x/1-x²)
(iv) tan-1(√(1+x²) – 1/x) w.r.t tan-1 x.
Sol: (i) Given y = tan-1(2x/1-x²) … (1)
and z = tan-1x … (2)
So we want to diff. y w.r.t. z i.e. to find dy/dz
put x = tan θ ⇒ θ = tan-1x
(ii) Given y =tan-1(x/√1-x²)… (1)
and z =sec-1(1/2x²-1)… (2)
So we want to diff. y w.r.t. z i.e. to find dy/dz
putting x = sin θ in eqn. (1); we get
(iii) Given y =tan-1(2x/1-x²)… (1)
and z =sin-1(2x/1-x²)… (2)
So we want to diff. y w.r.t. z i.e. to find dy/dz
putting x = tan θ ⇒ θ = tan-1x in eqn. (1); we have
(iv) Given y =tan-1(√(1+x²) – 1/x)… (1)
and z = tan-1 x … (2)
So we want to diff. y w.r.t. z i.e. to find dy/dz
putting x = tan θ ⇒ θ = tan-1x in eqn. (1); we have
–: End of Differentiation Class 12 OP Malhotra Exe-8J ISC Math Ch-8 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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