Home/Boards/Maharashtra Msbshse/Class 12 Science/Mathematics/Ch 8 — Differentiation/Find (dy)/(dx) if x=acotθ, y=bcosecθ.EXERCISE 1.4 · Q152Q.Find dydx\dfrac{dy}{dx}dxdy if x=acotθx=a\cot\thetax=acotθ, y=b cosec θy=b\,\text{cosec}\,\thetay=bcosecθ.Maharashtra MsbshseTextbookSubjectiveImportance★★★★★52% · 152/293 QuestionsConceptShort answerLong answerMethodsCommon mistakesRelatedPYQ – Same Concept✓ Free questionConcept understanding — Derivatives of Parametric FunctionsSometimes both xxx and yyy are given not directly in terms of each other but through a third variable ttt (the parameter): x=f(t)x=f(t)x=f(t), y=g(t)y=g(t)y=g(t). For example x=acostx=a\cos tx=acost, y=asinty=a\sin ty=asint traces a circle x2+y2=a2x^2+y^2=a^2x2+y2=a2 as ttt varies, without yyy ever being written explicitly as a function of xxx. When xxx and yyy are both differentiable functions of the parameter ttt with dxdt≠0\dfrac{dx}{dt}\ne0dtdx=0, then yyy is a differentiable function of xxx, and dydx=dy/dtdx/dt\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}dxdy=dx/dtdy/dt — found by differentiating yyy and xxx separately with respect to ttt and dividing. This follows from the increment definition: δyδx=δy/δtδx/δt\dfrac{\delta y}{\delta x}=\dfrac{\delta y/\delta t}{\delta x/\delta t}δxδy=δx/δtδy/δt, and taking the limit as δt→0\delta t\to0δt→0 (so δx→0\delta x\to0δx→0 too, since xxx is continuous in ttt) gives the result. To evaluate dy/dxdy/dxdy/dx at a particular value of the parameter, compute dy/dtdy/dtdy/dt and dx/dtdx/dtdx/dt as functions of ttt, form the ratio, and then substitute the given value of ttt (or θ\thetaθ) at the end. Differentiate w.r.t. θ\thetaθ, simplify cotθ/cosec θ=cosθ\cot\theta/\text{cosec}\,\theta=\cos\thetacotθ/cosecθ=cosθ. ✓Final answer dydx=bacosθ\dfrac{dy}{dx}=\dfrac{b}{a}\cos\thetadxdy=abcosθ We have x=acotθx=a\cot\thetax=acotθ and y=b cosec θy=b\,\text{cosec}\,\thetay=bcosecθ. Step 1. dxdθ=−a cosec2θ\dfrac{dx}{d\theta}=-a\,\text{cosec}^2\thetadθdx=−acosec2θ Step 2. dydθ=−b cosec θcotθ\dfrac{dy}{d\theta}=-b\,\text{cosec}\,\theta\cot\thetadθdy=−bcosecθcotθ Step 3. dydx=−b cosec θcotθ−a cosec2θ=bcotθa cosec θ\frac{dy}{dx}=\frac{-b\,\text{cosec}\,\theta\cot\theta}{-a\,\text{cosec}^2\theta}=\frac{b\cot\theta}{a\,\text{cosec}\,\theta}dxdy=−acosec2θ−bcosecθcotθ=acosecθbcotθ Step 4. Since cotθcosec θ=cosθ/sinθ1/sinθ=cosθ\dfrac{\cot\theta}{\text{cosec}\,\theta}=\dfrac{\cos\theta/\sin\theta}{1/\sin\theta}=\cos\thetacosecθcotθ=1/sinθcosθ/sinθ=cosθ, dydx=bacosθ\frac{dy}{dx}=\frac{b}{a}\cos\thetadxdy=abcosθ ✓Final answer dydx=bacosθ\dfrac{dy}{dx}=\dfrac{b}{a}\cos\thetadxdy=abcosθ Parametric differentiation using derivatives of cot and cosec, then trig simplification.Sign errors: forgetting that both d(cotθ)/dθ and d(cscθ)/dθ carry a negative sign (the negatives cancel, they don't add up).Maharashtra HSC (MSBSHSE) Board 2025If x=f(t)x=f(t)x=f(t) and y=g(t)y=g(t)y=g(t) are differentiable functions of ttt so that yyy is a function of xxx and if dxdt≠0\dfrac{dx}{dt}\ne 0dtdx=0 then prove that dydx=dy/dtdx/dt\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}dxdy=dx/dtdy/dt. Hence find the derivatView this question →Maharashtra HSC (MSBSHSE) Board 2024The slope of the tangent to the curve x=sinθx=\sin\thetax=sinθ and y=cos2θy=\cos 2\thetay=cos2θ at θ=π6\theta = \dfrac{\pi}{6}θ=6π is ____.View this question →Maharashtra HSC (MSBSHSE) Board 2024If x=f(t)x=f(t)x=f(t) and y=g(t)y=g(t)y=g(t) are differentiable functions of ttt, so that yyy is function of xxx and dxdt≠0\dfrac{dx}{dt}\ne 0dtdx=0 then prove that dydx=dy/dtdx/dt\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}dxdy=dx/dtdy/dt. Hence find $\dfrac{dy}{dx}View this question →Maharashtra HSC (MSBSHSE) Board 2022If x=f(t)x = f(t)x=f(t) and y=g(t)y = g(t)y=g(t) are differentiable functions of ttt so that yyy is differentiable function of xxx and dxdt≠0\dfrac{dx}{dt} \ne 0dtdx=0, then prove that: dydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}dxdy=dx/dtdy/dt. HenView this question →Maharashtra HSC (MSBSHSE) Board 2019If x=f(t)x = f(t)x=f(t) and y=g(t)y = g(t)y=g(t) are differentiable functions of ttt, then prove that yyy is a differentiable function of xxx andView this question →Maharashtra HSC (MSBSHSE) Board 2018If x=acos3tx = a\cos^3 tx=acos3t, y=asin3ty = a\sin^3 ty=asin3t, show that dydx=−(yx)1/3\dfrac{dy}{dx} = -\left(\dfrac{y}{x}\right)^{1/3}dxdy=−(xy)1/3.View this question →Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.CBSE Board6 questionsCompetitive Exams0 questions2 marks (1)3 marks (1)4 marks (4)CBSE 2024Set ANNUAL2 marksMCQQ.The slope of the tangent to the curve x=sinθx=\sin\thetax=sinθ and y=cos2θy=\cos 2\thetay=cos2θ at θ=π6\theta = \dfrac{\pi}{6}θ=6π is ____.(a) −23-2\sqrt{3}−23(b) −23\dfrac{-2}{\sqrt{3}}3−2(c) −2-2−2(d) −12-\dfrac{1}{2}−21›Reveal solutionSolution dydx=dy/dθdx/dθ=−4sinθ\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=-4\sin\thetadxdy=dx/dθdy/dθ=−4sinθ. x=sinθ⇒dxdθ=cosθx=\sin\theta \Rightarrow \dfrac{dx}{d\theta}=\cos\thetax=sinθ⇒dθdx=cosθ y=cos2θ⇒dydθ=−2sin2θ=−4sinθcosθy=\cos2\theta \Rightarrow \dfrac{dy}{d\theta}=-2\sin2\theta=-4\sin\theta\cos\thetay=cos2θ⇒dθdy=−2sin2θ=−4sinθcosθ dydx=−4sinθcosθcosθ=−4sinθ\dfrac{dy}{dx}=\dfrac{-4\sin\theta\cos\theta}{\cos\theta}=-4\sin\thetadxdy=cosθ−4sinθcosθ=−4sinθ At θ=π6\theta=\dfrac{\pi}{6}θ=6π: dydx=−4×12=−2\dfrac{dy}{dx}=-4\times\dfrac12=-2dxdy=−4×21=−2 ✓Final answer (c) −2-2−2 🎓Unlock everything free for 14 days✓Full step-by-step solutions✓Concept-first explanations✓Methods, shortcuts & mistakes✓PYQ mapping + timed mock testsStart 14-day free trial →See Plans & Pricing →Full access for 14 days. 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