6 Most FAQ’s of Differentiation Chapter in Inter 1st Year Maths-1B (TS/AP)

7 Marks

LAQ-1 : If √1 – x² + √1 – y² = a(x – y) then prove that dy/dx = √1 – y²/√1 – x²

$$\text{Given } \sqrt{1 – x^2} + \sqrt{1 – y^2} = a(x – y)$$

$$\text{We take } x = \sin \alpha, y = \sin \beta \text{ then}$$

$$\sqrt{1 – \sin^2 \alpha} + \sqrt{1 – \sin^2 \beta} = a(\sin \alpha – \sin \beta)$$

$$\Rightarrow \cos \alpha + \cos \beta = a(\sin \alpha – \sin \beta)$$

$$\Rightarrow \frac{\cos \alpha + \cos \beta}{\sin \alpha – \sin \beta} = a$$

$$\Rightarrow \frac{2\cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha – \beta}{2}\right)}{2\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha – \beta}{2}\right)} = a$$

$$\Rightarrow \frac{\cos\left(\frac{\alpha – \beta}{2}\right)}{\sin\left(\frac{\alpha – \beta}{2}\right)} = a$$

$$\Rightarrow \cot\left(\frac{\alpha – \beta}{2}\right) = a$$

$$\Rightarrow \frac{\alpha – \beta}{2} = \cot^{-1}(a)$$

$$\Rightarrow \alpha – \beta = 2\cot^{-1}(a)$$

$$\text{But } \sin \alpha = x$$

$$\Rightarrow \alpha = \sin^{-1}x \text{ and } y = \sin \beta \Rightarrow \beta = \sin^{-1}y$$

$$\sin^{-1}x – \sin^{-1}y = 2\cot^{-1}(a)$$

$$\text{On differentiating w.r.to } x \text{ we get}$$

$$\frac{1}{\sqrt{1 – x^2}} – \frac{1}{\sqrt{1 – y^2}} \frac{dy}{dx} = 0$$

$$\Rightarrow \frac{1}{\sqrt{1 – y^2}} \frac{dy}{dx} = \frac{1}{\sqrt{1 – x^2}}$$

$$\Rightarrow \frac{dy}{dx} = \frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}$$

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LAQ-2 : If y = Tan-1 (√1 + x² + √1 – x²/√1 + x² – √1 – x²) then find d dy/dx

$$\text{Given } y = \tan^{-1} \left(\frac{\sqrt{1 + x^2} + \sqrt{1 – x^2}}{\sqrt{1 + x^2} – \sqrt{1 – x^2}}\right)$$

$$\text{We take } x^2 = \cos 2\theta, \text{ then}$$

$$y = \tan^{-1} \left(\frac{\sqrt{1 + \cos 2\theta} + \sqrt{1 – \cos 2\theta}}{\sqrt{1 + \cos 2\theta} – \sqrt{1 – \cos 2\theta}}\right)$$
$$= \tan^{-1}\left(\frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta – \sqrt{2}\sin\theta}\right)$$

$$= \tan^{-1}\left(\frac{\sqrt{2} [\cos \theta + \sin \theta]}{\sqrt{2} [\cos \theta – \sin \theta]}\right)
= \tan^{-1} \left(\frac{\cos \theta + \sin \theta}{\cos \theta – \sin \theta}\right)$$

$$= \tan^{-1} \left(\frac{\cos \theta / \cos \theta + \sin \theta / \cos \theta}{\cos \theta / \cos \theta – \sin \theta / \cos \theta}\right)$$
$$= \tan^{-1} \left(\frac{1 + \tan \theta}{1 – \tan \theta}\right)$$

$$= \tan^{-1} [\tan(\pi/4 + \theta)]$$
$$= \pi/4 + \theta$$

$$y = \pi/4 + \theta$$

$$\text{But } \cos 2\theta = x^2$$

$$\Rightarrow 2\theta = \cos^{-1}(x^2)$$

$$\Rightarrow \theta = \frac{1}{2} \cos^{-1}(x^2)$$

$$\text{So, } y = \pi/4 + \frac{1}{2} \cos^{-1}(x^2)$$

$$\text{On differentiating w.r.t } x \text{ we get}$$

$$dy/dx = 0 + \frac{1}{2} \left(-\frac{1}{\sqrt{1 – (x^2)^2}} \cdot 2x\right)$$

$$dy/dx = -\frac{x}{\sqrt{1 – x^4}}$$

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LAQ-3 : Find the derivative of x^tanx + (sinx)^cosx w.r.to x

$$\text{Let } y = x \tan x + (\sin x)\cos x \quad (1)$$

$$\text{Take } u = x \tan x$$

$$\text{Taking log on both sides, we get}$$

$$\log u = \log(x \tan x)$$

$$\Rightarrow \log u = \tan x (\log x)$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{u} \frac{du}{dx} = \tan x \left(\frac{1}{x}\right) + \log x (\sec^2 x)$$

$$\Rightarrow \frac{1}{u} \frac{du}{dx} = \frac{\tan x}{x} + \log x \sec^2 x$$

$$\Rightarrow \frac{du}{dx} = u \left(\frac{\tan x}{x} + \log x \sec^2 x\right)$$

$$\Rightarrow \frac{du}{dx} = x \tan x \left(\frac{\tan x}{x} + \log x \sec^2 x\right)$$

$$\text{Take } v = (\sin x) \cos x$$

$$\text{Taking log on both sides, we get}$$

$$\log v = \log((\sin x) \cos x)$$

$$\Rightarrow \log v = \cos x[\log(\sin x)]$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{v} \frac{dv}{dx} = \cos x \left(\frac{1}{\sin x} \cos x\right) + \log(\sin x)(-\sin x)$$

$$\Rightarrow \frac{1}{v} \frac{dv}{dx} = \cos x \cot x – \log(\sin x) \sin x$$

$$\Rightarrow \frac{dv}{dx} = v \left(\cos x \cot x – \sin x \log(\sin x)\right)$$

$$\Rightarrow \frac{dv}{dx} = (\sin x) \cos x \left(\cos x \cot x – \sin x \log(\sin x)\right)$$

$$\text{From (1) }$$

$$y = u + v$$

$$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$

$$\frac{dy}{dx} = x \tan x \left(\frac{\tan x}{x} + \log x \sec^2 x\right) + (\sin x) \cos x \left(\cos x \cot x – \sin x \log(\sin x)\right)$$

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LAQ-4 : Find the derivative of (sinx)^logx + x^sinx

$$\text{Let } y = (\sin x)\log x$$

$$\text{Taking log on both sides, we get}$$

$$\log u = \log((\sin x)\log x)$$

$$\Rightarrow \log u = \log x [\log(\sin x)]$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{u} \frac{du}{dx} = \log x \left(\frac{\cos x}{\sin x}\right) + \log(\sin x) \left(\frac{1}{x}\right)$$

$$\Rightarrow \frac{1}{u} \frac{du}{dx} = \log x \cot x + \frac{\log(\sin x)}{x}$$

$$\Rightarrow \frac{du}{dx} = u(\log x \cot x + \frac{\log(\sin x)}{x})$$

$$\Rightarrow \frac{du}{dx} = (\sin x \log x)(\log x \cot x + \frac{\log(\sin x)}{x})$$

$$\text{Take } v = x \sin x$$

$$\text{Taking log on both sides, we get}$$

$$\log v = \log(x \sin x)$$

$$\Rightarrow \log v = \sin x [\log x]$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{v} \frac{dv}{dx} = \sin x \left(\frac{1}{x}\right) + \log x \cos x$$

$$\Rightarrow \frac{1}{v} \frac{dv}{dx} = \frac{\sin x}{x} + \log x \cos x$$

$$\Rightarrow \frac{dv}{dx} = v(\frac{\sin x}{x} + \log x \cos x)$$

$$\Rightarrow \frac{dv}{dx} = x \sin x \left(\frac{\sin x}{x} + \log x \cos x\right)$$

$$\text{From (1) }$$

$$y = u + v$$

$$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$

$$\frac{dy}{dx} = (\sin x \log x) (\log x \cot x + \frac{\log(\sin x)}{x}) + x \sin x (\frac{\sin x}{x} + \log x \cos x)$$

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LAQ-5 : If x^y + y^x = ab then show that dy/dx = -(yx^y-1 + y^xlogy/x^ylogx + xy^x-1)

$$\text{Given } xy + y^x = ab \quad (1)$$

$$\text{Take } u = xy$$

$$\text{Taking log on both sides, we get}$$

$$\log u = \log(xy)$$

$$\Rightarrow \log u = y \log x$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{u} \frac{du}{dx} = y \left(\frac{1}{x}\right) + \log x \frac{dy}{dx}$$

$$\Rightarrow \frac{1}{u} \frac{du}{dx} = \frac{y}{x} + \log x \frac{dy}{dx}$$

$$\Rightarrow \frac{du}{dx} = u \left(\frac{y}{x} + \log x \frac{dy}{dx}\right)$$

$$\Rightarrow \frac{du}{dx} = xy \left(\frac{y}{x} + \log x \frac{dy}{dx}\right)$$

$$\text{Take } v = y^x$$

$$\text{Taking log on both sides, we get}$$

$$\log v = \log(y^x)$$

$$\Rightarrow \log v = x \log y$$

$$\text{On differentiating using the product rule, we get}$$

$$\frac{1}{v} \frac{dv}{dx} = x \left(\frac{1}{y} \right) \frac{dy}{dx} + \log y$$

$$\Rightarrow \frac{1}{v} \frac{dv}{dx} = \frac{x}{y} \frac{dy}{dx} + \log y$$

$$\Rightarrow \frac{dv}{dx} = v \left(\frac{x}{y} \frac{dy}{dx} + \log y\right)$$

$$\Rightarrow \frac{dv}{dx} = y^x \left(\frac{x}{y} \frac{dy}{dx} + \log y\right)$$

$$\text{From (1) }$$

$$u + v = ab$$

$$\Rightarrow \frac{du}{dx} + \frac{dv}{dx} = 0$$

$$\Rightarrow xy \left(\frac{y}{x} + \log x \frac{dy}{dx}\right) + y^x \left(\frac{x}{y} \frac{dy}{dx} + \log y\right) = 0$$

$$\Rightarrow \left(xy \frac{y}{x} + xy \log x \frac{dy}{dx}\right) + \left(y^x \frac{x}{y} \frac{dy}{dx} + y^x \log y\right) = 0$$

$$\Rightarrow \frac{dy}{dx} \left(xy \log x + y^x \frac{x}{y}\right) + \left(xy \frac{y}{x} + y^x \log y\right) = 0$$

$$\Rightarrow \frac{dy}{dx} \left(xy \log x + y^x \frac{x}{y}\right) = -\left(xy \frac{y}{x} + y^x \log y\right)$$

$$\Rightarrow \frac{dy}{dx} = -\frac{xy \frac{y}{x} + y^x \log y}{xy \log x + y^x \frac{x}{y}}$$

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LAQ-6 : If y = x√a² + x² + a² log(x + √a² + x²), then show that dy/dx = 2√a² + x²

$$y = x \sqrt{a^2 + x^2} + a^2 \log(x + \sqrt{a^2 + x^2})$$

$$\frac{dy}{dx} = \left[ x \frac{d}{dx} \sqrt{a^2 + x^2} + \sqrt{a^2 + x^2} \frac{d}{dx} (x) \right] + a^2 \left[ \frac{d}{dx} \log(x + \sqrt{a^2 + x^2}) \right]$$

$$\frac{dy}{dx} = \left( x \frac{1}{2\sqrt{a^2 + x^2}} \frac{d}{dx} (a^2 + x^2) + \sqrt{a^2 + x^2} \cdot 1 \right) + a^2 \left( \frac{1}{x + \sqrt{a^2 + x^2}} \frac{d}{dx} (x + \sqrt{a^2 + x^2}) \right)$$

$$\frac{dy}{dx} = \left( \frac{x}{2\sqrt{a^2 + x^2}} \cdot 2x + \sqrt{a^2 + x^2} \right) + a^2 \left( \frac{1}{x + \sqrt{a^2 + x^2}} \cdot \left(1 + \frac{x}{2\sqrt{a^2 + x^2}} \cdot 2x \right) \right)$$

$$\frac{dy}{dx} = \left( \frac{x^2}{\sqrt{a^2 + x^2}} + \sqrt{a^2 + x^2} \right) + \left( \frac{a^2}{x + \sqrt{a^2 + x^2}} \cdot \left(\sqrt{a^2 + x^2} + \frac{x^2}{\sqrt{a^2 + x^2}} \right) \right)$$

$$\frac{dy}{dx} = \left( \frac{x^2 + a^2 + x^2}{\sqrt{a^2 + x^2}} \right) + \left( \frac{a^2 \left(\sqrt{a^2 + x^2} + \frac{x^2}{\sqrt{a^2 + x^2}}\right)}{x + \sqrt{a^2 + x^2}} \right)$$

$$\frac{dy}{dx} = \frac{a^2 + 2x^2}{\sqrt{a^2 + x^2}} + \frac{a^2 (\sqrt{a^2 + x^2} + x^2 / \sqrt{a^2 + x^2})}{x + \sqrt{a^2 + x^2}}$$

$$\frac{dy}{dx} = 2 \sqrt{a^2 + x^2}$$