Monthly Archives: March 2021

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

Annual Essay Contest: Mentoris Project: Aug 2021

https://mentorisproject.org/essay-contest

Match making, STEM, math, algorithm by 18 year old

https://www.npr.org/2021/03/18/978721494/matchmaker-matchmaker-make-me-an-algorithm-stem-contest-winner-pairs-data

Derivatives: part 11: IITJEE maths tutorial problems for practice

Problem 1: Find \frac{d}{dx}\arctan{\frac{4x}{4-x^{2}}}.

Choose (a) \frac{1}{4-x^{2}} (b) \frac{1}{4+x^{2}} (c) \frac{4}{4+x^{2}} (d) \frac{4}{4-x^{2}}

Solution 1:

Let y = \arctan{\frac{4x}{4-x^{2}}}. Hence, \tan{y} = \frac{4x}{4-x^{2}}. Differentiating both sides w.r.t. x, we get the following:

\sec^{2}{y} \times \frac{dy}{dx}= \frac{d}{dx} (\frac{4x}{4-x^{2}})

\sec^{2}{y} \times \frac{dy}{dx} = \frac{(4-x^{2}) \times 4 - 4x \times (-2x)}{(4-x^{2})^{2}} = \frac{16+4x^{2}}{(4-x^{2})}

But, \sec^{2}{y}=\tan^{2}{y}+1=\frac{(x^{2}+4)^{2}}{(4-x^{2})^{2}}

Hence, the answer is \frac{dy}{dx}= \frac{4}{4+x^{2}}. Option c.

Problem 2: Find \frac{dy}{dx} if \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}=1

Choose (a) \frac{-(2x+y)}{x+y} (b) \frac{-(2x+y)}{x+2y} (c) \frac{x+2y}{x+y} (d) -\frac{2x+y}{x+2y}

Solution 2:

The given equation is x+y = \sqrt{xy}. Differentiating both sides wrt x,

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

1+\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{\sqrt{x}}{2\sqrt{y}} \times \frac{dy}{dx}

(1- \frac{1}{2}\sqrt{\frac{x}{y}}) \times \frac{dy}{dx} =\frac{\sqrt{y}}{2\sqrt{x}} -1

\frac{dy}{dx} = \frac{\sqrt{y}-2\sqrt{x}}{2\sqrt{y}-\sqrt{x}} \times \frac{2\sqrt{y}}{2\sqrt{x}}

\frac{}{} = \frac{2y - 4\sqrt{xy}}{4\sqrt{xy}-2x} = \frac{2y-4(x+y)}{4(x+y)-2x} = - \frac{2x+y}{x+2y} is the answer. Option D.

Problem 3: If y=\arctan{\frac{\log{(\frac{e}{x^{2}})}}{\log{(ex^{2})}}} then \frac{dy}{dx} is

choose (a) e (b) \frac{2}{x(1+4(\log{x})^{2})}(c) \frac{-2}{x(1+4(\log{x})^{2})} (d) \frac{2}{1+x^{2}}

Solution 3:

Given that y = \arctan{(\frac{\log(\frac{e}{x^{2}})}{\log(ex^{2})})} so that we have

\tan{y} = \frac{\log{\frac{e}{x^{2}}}}{\log{ex^{2}}} so now differentiating both sides w.r.t. x,

\sec^{2}{y}\frac{dy}{dx} = \frac{\frac{\log{(ex^{2})}}{\frac{e}{x^{2}}} \frac{d}{dx}(\frac{e}{x^{2}}) - \log{(\frac{e}{x^{2}})} \times \frac{1}{ex^{2}} \times \frac{d}{dx}(ex^{2})}{(\log{(ex^{2})})}

\sec^{2}{y}\frac{dy}{dx} = \frac{-\frac{2}{x}(\log{(ex^{2})})- \frac{2}{x}\log{(\frac{e}{x^{2}})}}{(\log{(ex^{2})})^{2}}

\sec^{y}(\frac{dy}{dx}) = \frac{-\frac{2}{x}(\log{(ex^{2}) \times (\frac{e}{x^{2}})})}{(\log{ex^{2}})^{2}}

\sec^{y}\frac{dy}{dx} = \frac{-\frac{4}{x}}{(\log{(ex^{2})^{2}})} = \frac{-4}{x(\log{(ex^{2})})^{2}}

Now, we also know that\sec^{2}{y} = 1 + \tan^{2}{y} = \frac{(\log{(\frac{e}{x^{2}})})^{2}}{(\log{(ex^{2})})^{2}} + 1 = \frac{(\log(\frac{e}{x^{2}}))^{2}+(\log{(ex^{2})})^{2}}{(\log{(ex^{2})})^{2}}

But, note that by laws of logarithms, on simplification, we get

\log{(\frac{e}{x^{2}})} = 1 - 2\log{x} and \log{(ex^{2})} = 1 + 2 \log{x} so that on squaring, we get

(\log{(e/x^{2})})^{2} = 1-4\log{x} + 4 (\log{x})^{2}

(\log{(ex^{2})})^{2}=1+4\log{x} + 4 (\log{x})^{2} so that now we get

(\log{(\frac{e}{x^{2}})})^{2} + (\log{(ex^{2})})^{2} = 2 + 8 (\log{x})^{2}, which all put together simplifies to

\frac{dy}{dx} = \frac{1}{\sec^{2}{y}} \times \frac{-(\frac{4}{x})}{(\log{(ex^{2})})^{2}}

\frac{dy}{dx} = - \frac{(\frac{2}{x})}{1+4(\log{x})^{2}} so that the answer is option C.

Problem 4: Find \frac{d}{dx}(\arcsin{(3x-4x^{3})}+\arccos{(2x(\sqrt{(1-x^{2})}))})

Choose option (a) \frac{1}{\sqrt{1-x^{2}}} (b) \frac{-1}{\sqrt{1-x^{2}}} (c) \frac{5}{\sqrt{1-x^{2}}} (d) \frac{-2}{\sqrt{1-x^{2}}}

Solution 4:

Let us consider the first differential. Let us substitute x = \sin{\theta}. Hence,

3x-4x^{3}=3\sin{\theta} - 4\sin^{3}{\theta}= \sin{3\theta} and so we \arcsin{3x-4x^{3}} = \arcsin{\sin{3\theta}} = 3 \theta, and so also, we get \arccos{2x\sqrt{1-x^{2}}}=\arccos{2\sin{\theta}\cos{\theta}} = \arccos{\cos{2\theta}}=2\theta so we get

required derivative

\frac{dy}{dx} = \frac{d}{dx}(3\theta) + \frac{d}{dx}(2\theta) = \frac{d}{dx}(5\theta) = 5 \frac{d\theta}{dx} = 5 \frac{d}{dx}(\arcsin{x})= 5 \frac{1}{\sqrt{1-x^{2}}}. Answer is option C.

Problem 5: Find \frac{d}{dx}(x-a)(x-b)(x-c)\ldots (x-z)

Choose option (a) zero (b) 26 (c) 26! (d) does not exist

Solution 5: the expression also includes a term 0 = (x-x) so that the final answer is zero only.

Problem 6: Find \frac{d}{dx}(x^{x})^{x}.

Solution 6: Let y= (x^{x})^{x}

so \log{y} = \log{(x^{x})^{x}}

so \log{y} = x^{2} \times \log{x} so that differentiating both sides w.r.t. x, we get

we get \frac{1}{y} \times \frac{dy}{dx} = \frac{x^{2}}{x} + \log{x} \times 2x

we get \frac{1}{y} \times \frac{dy}{dx} = x + 2x \log{x} = x(1+2\log{x})

we get \frac{dy}{dx} = yx (1+2 \log{x}) = (x^{x})^{x} \times x \times (1+2\log{x})

so the answer is option B.

Choose option (a): x.x^{x}(1+2\log{x}) (b) x^{x^{2}+1} \times (1+2\log{x}) (c) {x^{{x}^{2}}}(1+\log{x}) (d) none of these

Problem 7:

Find \frac{d}{dx}(e^{x^{x}})

Choose option (a) e^{x^{x}}.x^{x}.(1+\log{x}) (b) e^{x^{x}}. x^{x}.\log{(\frac{x}{e})} (c) e^{x^{x}}.x^{x} (d) e^{x^{x}}. (\log{(e^{x})})

Solution 7: Let y = (e^{(x^{x})}) so that taking logarithm of both sides

\log{y} = \log{(e^{(x^{x})})} so that \log{y} = x^{x} \log{e} = x^{x}

\log {(\log{y})}= x \times (\log{x}). Differentiating both sides w.r.t.x we get:

\frac{1}{\log{y}} \times \frac{d}{dx} \times (\log{y})= \frac{x}{x} + \log{x} so that we get now

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

\frac{dy}{dx} = e^{x^{x}} \times x^{x} \times (1+\log{x}) so we get option a as the answer.

Problem 8:

Find \frac{d}{dx}(x^{x^{x}})

Choose option (a): x^{x^{x}} \times (1+\log{x}) (b) x^{x^{x}} \times (x^{x}\log{x})(1+\log{x}+\frac{1}{x}) (c) x^{x^{x}} \times (x^{x}\log{x}) \times (1+\log{x}+\frac{1}{x\log{x}}) (d) none of these.

Solution 8:

let y=x^{x^{x}} taking logarithm of both sides we get

\log{y} = x^{x} \times \log{x} and now differentiating both sides w.r.t.x, we get

\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{x}}{x} + (\log{x}) \times \frac{d}{dx} (x^{x}) and now let t=x^{x} and again take logarithm of both sides so that we get (this is quite a classic example…worth memorizing and applying wherever it arises):

\log{t}= x\log{x}

\frac{1}{t} \frac{dt}{dx} = \frac{x}{x} + \log{x}=1+\log{x}

\frac{dt}{dx} = x^{x}(1+\log{x})

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

\frac{dy}{dx} = x^{x}(x^{x-1}+x^{x} \times \log{x} \times (1+\log{x}))

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

The answer is option C.

Problem 9:

Find \frac{d}{dx}(x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8}).

Choose option (a): \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}} (b) \frac{x^{16}-a^{16}}{x-a} (c) \frac{x^{16}-x^{15}a+a^{16}}{(x-a)^{2}} (d) none of these

Solution 9:

Given that y = (x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})

Remark: Simply multpilying out thinking the symmetry will simplify itself is going to lead to a mess…because there will be no cancellation of terms …:-) The way out is a simple algebra observation…this is why we should never ever forget the fundamentals of our foundation math:-)

note that the above can be re written as follows:

y = \frac{(x^{2}-a^{2})}{(x-a)} \times \frac{(x^{4}-a^{4})}{(x^{2}-a^{2})} \times \frac{x^{8}-a^{8}}{(x^{4}-a^{4})} \times \frac{(x^{16}-a^{16})}{(x^{8}+a^{8})}

Now, we are happy like little children because many terms cancel out 🙂 hahaha…lol 🙂

y = \frac{(x^{16}-a^{16})}{(x-a)} and now differentiating both sides w.r.t.x we get

\frac{dy}{dx} = \frac{(x-a)(16x^{15})- (x^{16}-a^{16})(1)}{(x-a)^{2}}

\frac{dy}{dx} = \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}

The answer is option A.

Problem 10:

If x= \theta {\cos{\theta}}+\sin{\theta} and y = \cos{\theta}-\theta \times \sin{\theta} then find the value of \frac{dy}{dx} at\theta = \frac{\pi}{2}

Choose option (a): -\frac{\pi}{2} (b) \frac{2}{\pi} (c) \frac{\pi}{4} (d) \frac{4}{\pi}

Solution 10:

\frac{dx}{d\theta} = \cos{\theta} - \theta \times \sin{\theta} + \cos{\theta}

\frac{dy}{d\theta} = -\sin{\theta} - (\sin{\theta} + \theta \times \cos{\theta})

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

\frac{dx}{d\theta} = 2 \cos{\theta} - \theta \times \sin{\theta}

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dy}{d\theta}} = \frac{\theta \times \cos{\theta}-2\sin{\theta}}{2\cos{\theta}-\theta \times \sin{\theta}} = \frac{-2}{-\pi/2}=\frac{4}{\pi}

Answer is option D.

Regards,

Nalin Pithwa.