Derivatives: part 12:IITJEE maths tutorial problems for practice

1, x = a(t+\frac{1}{t}), y=a(t-\frac{1}{t}), then find \frac{dy}{dx}.

Option (A) \frac{t^{2}-1}{t^{2}+1}

Option (B) \frac{t^{2}+1}{t^{2}-1}

Option (C) \frac{t^{2}+1}{1-t^{2}}

Option (D) \frac{1}{t}

Solution 1: Given that x=at+\frac{a}{t} so that \frac{dx}{dt} = a +\frac{a(-1)}{t^{2}} = a(1-\frac{1}{t^{2}})

and given that y = at - \frac{a}{t} so that \frac{dy}{dt} = x + \frac{a}{t^{2}} = a(1+ \frac{1}{t^{2}})

and so we get \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(1 - \frac{1}{t^{2}})}{a(1+\frac{1}{t^{2}})} = \frac{t^{2}-1}{t^{2}+1} so that correct choice is option A.

2. If x = a \sin{3t} + b \cos{3t} and y= b \cos {t} + a \sin{t} then find \frac{dy}{dx} when t = \frac{\pi}{4}

Option (A) 0

Option (B) \frac{b-a}{3(a+b)}

Option (C) \frac{a-b}{3(a+b)}

Option (D) \frac{b-a}{b+a}

Solution 2:

\frac{dy}{dt} = -b \sin{t} + a\cos{t}

\frac{dx}{dt} = 3a \cos{3t} + -3b \sin{3t}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{-b\sin{t}+a \cos{t}}{3a \cos{3t}-3b\sin{3t}} which is equal to the following at t = \frac{\pi}{4}

\frac{dy}{dx} = \frac{- \frac{b}{\sqrt{2}} + \frac{a}{\sqrt{2}}}{-\frac{3a}{\sqrt{2}} - \frac{3b}{\sqrt{2}}}=-\frac{1}{3} \times \frac{b-a}{a+b} so that the correct choice is C.

3. If y = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} + \log{(\sqrt{1-x^{2}})}, then find \frac{dy}{dx}

Option A: \frac{\arcsin{x}}{(1-x^{2})^{\frac{3}{2}}}

Option B: \frac{\arcsin{x}}{\sqrt{1-x^{2}}}

Option C:\frac{\arcsin{x}}{1-x^{2}}

Option D: (1-x^{2})^{\frac{3}{2}}\arcsin{x}

Solution 3:

Let y = f(x) + g(x) where we put f(x) = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} so now let x=\sin{\theta}

So, we get \frac{dx}{d\theta} = \cos{\theta} and 1-x^{2}= \cos^{2}{\theta} and \sqrt{1-x^{2}} = \cos{\theta}

So we get f(\theta) = \frac{\theta \times \sin{\theta}}{\cos{\theta}} = \theta \times \tan{\theta}

So now \frac{df}{d\theta}= \tan{\theta}+ \theta \times \sin^{2}{\theta}

And, g(x) = \log{\sqrt{1-x^{2}}}

\frac{dg}{dx} =  \frac{1}{\sqrt{1-x^{2}}} \times \frac{d}{dx} (\sqrt{1-x^{2}}) = \frac{1}{2(1-x^{2})} \times (-2x)= \frac{-x}{1-x^{2}}

Hence, we get the following:

\frac{dy}{dx} = \frac{x}{\sqrt{1-x^{2}}}  + \frac{\arcsin{x}}{\frac{1}{1-x^{2}}} + \frac{-x}{1-x^{2}}

Question 4: Find the following: \frac{d}{dx}(sec^{-1}{(\frac{1}{\sqrt{1-x^{2}}})} + cot^{-1}(\frac{\sqrt{1-x^{2}}}{x}))

Option a: \frac{2}{\sqrt{1-x^{2}}}

Option b: \frac{1}{\sqrt{1-x^{2}}}

Option c: \frac{\sqrt{1-x^{2}}}{x}

Option d: \sqrt{1-x^{2}}

Solution 4:

Let f(x)=y_{1}=sec^{-1}(\frac{1}{\sqrt{1-x^{2}}})

Let x=\sin{\theta}, 1-x^{2}=\cos^{2}(\theta), \sqrt{1-x^{2}}=\cos{\theta}, and \frac{1}{\sqrt{1-x^{2}}} = \sec{\theta}

so y_{1}=\theta=\arcsin{x}

\frac{dy_{1}}{dx} = \frac{1}{\sqrt{1-x^{2}}}

Let y_{2}=cot^{-1}{\frac{\sqrt{1-x^{2}}}{x}}

Let x=\sin{\theta}, \sqrt{1-x^{2}}=\cos{\theta} and \frac{\sqrt{1-x^{2}}}{x}=\cot{\theta}

Let y_{2}= \cot^{-1}{\cot{\theta}}=\theta

so that \frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{x})=\frac{1}{1-x^{2}}

so that \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}} so the option is a.

Question 5:

If y=(x+\sqrt{1+x^{2}})^{n} then find (x^{2}+1)(\frac{dy}{dx})^{2}.

Solution 5:

y=(x+\sqrt{1+x^{2}})^{n}

\frac{dy}{dx} = x(x+\sqrt{1+x^{2}})^{n-1}\frac{d}{dx}(x+\sqrt{1+x^{2}}) = n(x+\sqrt{1+x^{2}})^{n-1}(1+\frac{2x}{2\sqrt{1+x^{2}}})

(\frac{dy}{dx})^{2}.(x^{2}+1) = (x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n-2} \times (1+\frac{x}{\sqrt{1+x^{2}}})^{2}

=(x^{2}+1).n^{2}. (x+\sqrt{1+x^{2}})^{2n-2} \times (\frac{x+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}})^{2}

=(x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n}.\frac{1}{(1+x^{2})}=n^{2}y^{2}

Question 6:

If f(x)= \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}, then f^{'}(0) is equal to

(a) 1/2 (b) 1/3 (c) 1/6 (d) 0

Solution 6:

Given that y = \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}

Hence, we have (x+3)(x+6) y^{2}=(x+1)(x+2)

(x+3)\frac{d}{dx}(y^{2}(x+6))+y^{2}(x+6) \times 1 = (x+1) \times 1 + (x+2) \times 1=2x+3

(x+3)(x+6)2y\frac{dy}{dx} + (x+3)y^{2} \times 1 =2x+3

(x+3)(x+6)2y\frac{dy}{dx} + (x+3) \times \frac{(x+1)(x+2)}{(x+3)(x+6)} = 2x+3

(x+3)(x+6)^{2}.2. \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}+(x+1)(x+2) = (2x+3)(x+6)

2\sqrt{(x+1)(x+2)(x+3)(x+6)} \times (x+6)\frac{dy}{dx} + (x+1)(x+2) = (2x+3)(x+6)

So, at x=0, on substitution we get f^{'}(0).

Question 7:

If y = \frac{1-t^{2}}{1+t^{2}}, x = \frac{2t}{1+t^{2}}, then find \frac{dy}{dx}.

Solution 7:

Given y= \frac{1-t^{2}}{1+t^{2}}, let t= \tan{\theta} so that \frac{dt}{d\theta}= \sec^{2}(\theta)

so that y = \frac{1-t^{2}}{1+t^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \frac{\cos^{2}{\theta}-\sin^{2}{\theta}}{1} = 2 \cos^{2}{\theta}-1= \cos{2\theta}

Now, x = \frac{2t}{1+t^{2}}=\frac{2\tan{\theta}}{1+\tan^{2}{\theta}} so that x = \sin{2\theta}

so now \frac{dx}{d\theta}=2 \cos{2\theta}

y = \cos{2\theta}

\frac{dy}{d\theta} = -2 \sin{2\theta}

\frac{dy}{dx} = - \frac{2\sin{(2\theta)}}{2(\cos{2\theta})}= - \tan{(2\theta)} = - \frac{2t}{1-t^{2}}= - \frac{x}{y}.

Question 8:

Find \frac{d}{dx}(\arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})})

Solution 8:

Let it be given that y = \arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}

Now, let us simplify this as y=y_{1}+y_{2} where y_{1} = \arctan{x} and y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}

Now, first consider y_{1} = \arctan{x}. Taking derivative of both sides w.r.t. x, we get

\frac{dy_{1}}{dx} = \frac{d}{dx}(\arctan{x}) = \frac{1}{1+x^{2}}….A

Now, next consider y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}. Takind derivative of both sides w.r.t. x, we get

\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{(\frac{x}{\sqrt{1+x^{2}}})}) = \frac{1}{1- \frac{x^{2}}{1+x^{2}}} \frac{d}{dx}(\frac{x}{\sqrt{1+x^{2}}}) = \frac{1}{1+x^{2}}(\frac{1}{\sqrt{1+x^{2}}+\frac{x}{2}(1+x^{2})^{-3/2}})….B

So that we get \frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}using A and B.

Question 9:

If x^{y} = y^{x}, then find \frac{dy}{dx}

Solution 9:

Given that x^{y} = y^{x}

y \log{x}= x \log{y}. Taking derivative of both sides w.r.t. x, we get

(\log{x}).\frac{dy}{dx} + \frac{y}{x} = \frac{x}{y}. \frac{dy}{dx} + (\log{y}) \times 1

(\log{x}- \frac{x}{y}).\frac{dy}{dx} = (\log{y}) - \frac{y}{x}

\frac{dy}{dx} = \frac{\frac{x(\log{y}-y)}{x}}{\frac{y\log{x}-x}{y}}= \frac{y}{x} \times \frac{x(\log{y})-y}{y(\log{x})-x} which is the required answer.

Question 10:

If (x+y)^{m+n} = x^{m}y^{n}, then find \frac{dy}{dx}.

Solution 10:

Given that (x+y)^{m+n} = x^{m}y^{n}

Taking logarithm of both sides w.r.t. any arbitrary valid base,

(m+n) \times \log{(x+y)} = \log{(x^{m}y^{n})} = \log(x^{m}) + \log{y^{n}} so that (m+n).\log{(x+y)}=m \log{x} + n \log{y}

Taking derivative of both sides w.r.t. x, we get the following:

\frac{m+n}{x+y}. \times (1+\frac{dy}{dx}) = \frac{m}{x} + \frac{n}{y}.\times \frac{dy}{dx}

\frac{m+n}{x+y} \times \frac{dy}{dx} - \frac{x}{y}. \frac{dy}{dx} = \frac{m}{x} - \frac{m+n}{x+y}

\frac{(m+n)y-n(x+y)}{y(x+y)}. \frac{}{} = \frac{mx+my-mx-nx}{x(x+y)}

\frac{my+ny-nx-ny}{} = \frac{my-nx}{x(x+y)}

\frac{my-nx}{y(n+y)}. \frac{dy}{dx} = \frac{my-nx}{x(x+y)}, so that finally we get the desired answer:

\frac{dy}{dx} = \frac{y}{x}

More later,

Cheers,

Nalin Pithwa

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