OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(h)

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S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(h)

Find \(\frac { dy }{ dx }\) if

Question 1.
x² + y² = a²
Solution:
Given x² + y² = a² ; Diff. both sides w.r.t. x, we have
2x + 2y \(\frac { dy }{ dx }\) = 0
⇒ \(\frac { dy }{ dx }\) = – \(\frac { x }{ y }\)

Question 2.
y² = 4ax
Solution:
Given y² = 4ax ; Diff. both sides w.r.t. x, we have
2y\(\frac { dy }{ dx }\) = 4a ⇒ \(\frac{d y}{d x}=\frac{2 a}{y}\)

Question 3.
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Solution:
Given \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (taking y as a function of x) ; Diff. both sides w.r.t. x, we have
\(\frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{b^2 x}{a^2 y}\)

Question 4.
xy = c²
Solution:
Given xy = c²; Diff. both sides w.r.t. x, we have
x\(\frac { dy }{ dx }\) + y = 0 ⇒ \(\frac { dy }{ dx }\) = – \(\frac { y }{ x }\)

Question 5.
x³ + 8xy + y³ = 64
Solution:
Given x³ + 8xy + y³ = 64
diff. both sides w.r.t. x, we have
3x² + 8\(\left[x \frac{d y}{d x}+y \cdot 1\right]+3 y^2 \frac{d y}{d x}\) = 0
⇒ (8x + 3y)²\(\frac { dy }{ dx }\) = – 3x² – 8y
∴ \(\frac { dy }{ dx }\) = – \(\frac{\left(3 x^2+8 y\right)}{8 x+3 y^2}\)

Question 6.
x³ + y³ = 3axy
Solution:
Given x³ + y³ = 3axy ; Diff. both sides w.r.t. x, we have
3x² + 3y²\(\frac{d y}{d x}=3 a\left[x \frac{d y}{d x}+y \cdot 1\right]\)
⇒ (3y² – 3ax)\(\frac { dy }{ dx }\) = 3ay – 3x²
∴ \(\frac{d y}{d x}=\frac{a y-x^2}{y^2-a x}\)

Question 7.
ax² + 2hxy + by² + 2gx + 2fy + c = 0
Solution:

Question 8.
(x² + y²)² = xy
Solution:
Given (x² + y²)² = xy ; Diff. both sides w.r.t. x, we have

Question 9.
\(\sqrt{x}+\sqrt{y}=\sqrt{a}\)
Solution:
Given \(\sqrt{x}+\sqrt{y}=\sqrt{a}\); Diff. both sides w.r.t. x, we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{-\sqrt{y}}{\sqrt{x}}=-\sqrt{\frac{y}{x}}\)

Question 10.
x² + y² = log (xy)
Solution:
Given x² + y² = log xy ; Diff. both sides w.r.t. x, we have

Question 11.
xⁿ + yⁿ = aⁿ
Solution:
Given xⁿ + yⁿ = aⁿ ; Diff. both sides w.r.t. x, we have
\(n x^{n-1}+n y^{n-1} \frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}=-\frac{x^{n-1}}{y^{n-1}}=-\left(\frac{x}{y}\right)^{n-1}\)

Question 12.
\(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}\)
Solution:
Given \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}\) ; Diff. both sides w.r.t. x, we have
\(\frac{2}{3} x^{\frac{1}{3}}+\frac{2}{3} y^{\frac{1}{3}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \Rightarrow \frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}\)

Question 13.
If y = x sin y, prove that x.
x\(\frac{d y}{d x}=\frac{y}{(1-x \cos y)}\).
Solution:

Question 14.
If ax² + 2hxy + by² = c , verify that \(\frac { dy }{ dx }\).\(\frac { dy }{ dx }\) = 1.
Solution:
Given ax² + 2hxy + by² = c² … (1)
Diff. eqn. (1) both sides w.r.t. x, we have

Question 15.
If sin^-1 \(\left(\frac{x^2-y^2}{x^2+y^2}\right)\) = c, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 16.
If y log x = x – y, prove that \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}\)
Solution:
Given y log x = x – y

Question 17.
If \(\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}\) = 6,
show that \(\frac{d y}{d x}=\frac{x-17 y}{17 x-y}\).
Solution:
Given \(\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}\) = 6 … (1)
on squaring eqn(1) both sides ; we have

Question 18.
Find \(\frac { dy }{ dx }\) 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
Solution:
(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 (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\(\frac { dy }{ dx }\) + x\(\frac { dy }{ dx }\)+y-1 = 0
⇒ (x – 2 sin y)\(\frac { dy }{ dx }\) = -(y + sin 2x)
⇒ \(\frac { dy }{ dx }\) = \(\frac{\sin 2 x+y}{2 \sin y-x}\)

Question 19.
If y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots . \text { to } \infty}}}\) prove that (2y – 1) \(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\).
Solution:
Given
y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots . \text { to } \infty}}}\)
⇒ y = \(\sqrt{\log x+y}\);
on squaring both sides;
we have y² = log x + y
DifF. both sides w.r.t. (x) ; we have
2y\(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\) + \(\frac { dy }{ dx }\)
⇒ (2y – 1) \(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\)

Question 20.
If y = \(\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \text { to } \infty}}}\), prove that (2y – 1) \(\frac { dy }{ dx }\) + sin x = 0.
Solution:
Given
y = \(\sqrt{\log x+y}\) ;
on squaring;
we have y² = cos x + y ;
DifF. both sides w.r.t. (x) ; we have
2y\(\frac { dy }{ dx }\) = – sin x + \(\frac { dy }{ dx }\)
⇒ (2y – 1) \(\frac { dy }{ dx }\) + sin x = 0