Differentiation Class 12 OP Malhotra Exe-8H ISC Maths Solutions Ch-8 Solutions. In this article you would learn about derivatives of implicit functions . 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-8H ISC Maths Solutions Ch-8
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-8 | Differentiation |
| Writer | OP Malhotra |
| Exe-8(H) | derivatives of implicit functions |
Derivatives of Implicit Functions
Differentiation Class 12 OP Malhotra Exe-8H ISC Maths Solutions Ch-8
Que-1: x² + y² = a²
Sol: Given x² + y² = a² ; Diff. both sides w.r.t. x, we have
2x + 2y dy/dx = 0
⇒ dy/dx = – x/y
Que-2: y² = 4ax
Sol: Given y² = 4ax ; Diff. both sides w.r.t. x, we have
2y dy/dx = 4a ⇒ dy/dx = 2a/y
Que-3: x²/a² + y²/b² = 1
Sol: Given x²/a² + y²/b² = 1 (taking y as a function of x) ; Diff. both sides w.r.t. x, we have
2x/a² + 2y/b² dy/dx = 0
⇒ dy/dx = -b²x/a²y
Que-4: xy = c²
Sol: Given xy = c²; Diff. both sides w.r.t. x, we have
x dy/dx + y = 0 ⇒ dy/dx = -y/x
Que-5: x³ + 8xy + y³ = 64
Sol: Given x³ + 8xy + y³ = 64
diff. both sides w.r.t. x, we have
3x² + 8[x dy/dx + y.1 ] +3y² dy/dx = 0
⇒ (8x + 3y)² dy/dx = -3x² – 8y
∴ dy/dx = – (3x²+8y)/8x+3y²
Que-6: x³ + y³ = 3axy
Sol: Given x³ + y³ = 3axy ; Diff. both sides w.r.t. x, we have
3x² + 3y² dy/dx = 3a [x dy/dx +y.1]
⇒ (3y² – 3ax) dy/dx = 3ay – 3x²
∴ dy/dx = ay-x²/y²-ax
Que-7: ax² + 2hxy + by² + 2gx + 2fy + c = 0
Sol:
Que-8: (x² + y²)² = xy
Sol: Given (x² + y²)² = xy ; Diff. both sides w.r.t. x, we have
Que-9: √x + √y = √a
Sol: Given √x + √y = √a ; Diff. both sides w.r.t. x, we have
1/2√x + 1/2√y dy/dx = 0
⇒ dy/dx = – √y/√x = -√(y/x)
Que-10: x² + y² = log (xy)
Sol: Given x² + y² = log xy ; Diff. both sides w.r.t. x, we have
Que-11: xn + yn = an
Sol: Given xn + yn = an ; Diff. both sides w.r.t. x, we have
n.(xn-1 ) + n.(yn-1 ) dy/dx = 0 ⇒ dy/dx = -(xn-1 )/(yn-1 ) = -(x/y)n-1
Que-12: x2/3 + y2/3 = a2/3
Sol: Given x2/3 + y2/3 = a2/3 ; Diff. both sides w.r.t. x, we have
2/3 (x1/3) + 2/3 (y1/3) dy/dx = 0
⇒ dy/dx = – (x-1/3)/(y-1/3) = -(y1/3)/(x1/3) ⇒ dy/dx = – (y/x)1/3
Que-13: If y = x sin y, prove that x.
x.dy/dx = y/(1-xcosy)
Sol:
Que-14: If ax² + 2hxy + by² = c , verify that (dy/dx).(dx/dy)=1
Sol: Given ax² + 2hxy + by² = c² … (1)
Diff. eqn. (1) both sides w.r.t. x, we have
Que-15: If sin-1 (x²-y²/x²+y²) = c, prove that dy/dx=y/x.
Sol:
Que-16: If y log x = x – y, prove that dy/dx=logx/(1+logx)²
Sol: Given y log x = x – y
Que-17: If √y/x + √x/y = 6 , show that dy/dx=x-17y/17x-y
Sol: Given √y/x + √x/y = 6 … (1)
on squaring eqn(1) both sides ; we have
Que-18: Find dy/dx if
(i) x = y log (xy)
(ii) x log y + y log x = 5
(iii) sin (x + y) = log (x + y)
(iv) sin²x + 2 cos y + xy = 0
Sol: (i) Given x = y log (xy)
⇒ x = y[log x + log y]
Diff. both sides w.r.t. x, we have
(ii) Given x log y + y log x = 5
Diff. both sides w.r.t. x, we have
(iii) Given sin (x + y) = log (x + y)
Diff. both sides w.r.t. x, we have
(iv) Given sin²x + 2 cos y + xy = 0
Diff. both sides w.r.t. x, we have
2 sin x cos x – 2 sin y dy/dx + x dy/dx + y-1 = 0
⇒ (x – 2 sin y) dy/dx = -(y + sin 2x)
⇒ dy/dx = sin 2x + y /2 sin y – x
Que-19: If y = √(logx + √logx +√logx + ……….. to ∞ ,prove that (2y-1) dy/dx = 1/x
Sol: Given
y =√(logx + √logx +√logx + ……….. to ∞
⇒ y = √(logx + y);
on squaring both sides;
we have y² = log x + y
Diff. both sides w.r.t. (x) ; we have
2y dy/dx =1/x + dy/dx
⇒ (2y – 1) dy/dx = 1/x
Que-20: If y = √(cosx + √cosx +√cosx + ……….. to ∞ , prove that (2y-1) dy/dx + sinx = 0
Sol: y = √logx + y ;
on squaring;
we have y² = cos x + y ;
Diff. both sides w.r.t. (x) ; we have
2y dy/dx = – sin x + dy/dx
⇒ (2y – 1) dy/dx + sin x = 0
–: End of Differentiation Class 12 OP Malhotra Exe-8H ISC Math Ch-8 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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