Derivative (1st order)

Study Material (PDF)
Self Assesment
Study Material (PDF)
Self Assesment

Click on each question to see answer & solution.

Question No: #81

If `x\sqrt{1+y}+y\sqrt{1+x}=0` then prove that `\frac{dy}{dx}=-\frac{1}{(1+x)^2}`

Question No: #82

If `\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)`, then show that `\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}`
Ans: N.A.
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Question No: #83

If `\sqrt{1+x^2}+\sqrt{1+y^2}=k(x-y)` where `k` constant, then show that `\frac{dy}{dx}=\sqrt{\frac{1+y^2}{1+x^2}}`

Question No: #84

If `y\sqrt{1-x^2}+x\sqrt{1-y^2}=1`, then show that `\frac{dy}{dx}=-\sqrt{\frac{1-y^2}{1-x^2}}`

Question No: #85

If `\sqrt{1-x^4}+\sqrt{1-y^4}=k(x^2-y^2)` where `k` constant, then show that `\frac{dy}{dx}=\frac{x\sqrt{1-y^4}}{y\sqrt{1-x^4}}`

Question No: #86

If `\sqrt{1-x^6}+\sqrt{1-y^6}=a^3(x^3-y^3)` where `a` is constant, then show that `\frac{dy}{dx}=\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}`

Question No: #87

If `xy=sin(x+y)` then show that `\frac{dy}{dx}=\frac{\sqrt{1-x^2y^2}-y}{x-\sqrt{1-x^2y^2}}`

Question No: #88

If `xy=tan(x+y)` then find `\frac{dy}{dx}`
Ans: `\frac{y-1-x^2y^2}{1-x+x^2y^2}`
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Question No: #89

If `xy=cot(x+y)` then show that `\frac{dy}{dx}=\frac{1+y+x^2y^2}{1+x+x^2y^2}`

Question No: #90

If `cosy=xcos(a+y)`, then prove that `\frac{dy}{dx}=\frac{cos^2(a+y)}{sina}` where `a\ne0` is a constant