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Continuity and Differentiability - Derivative of inverse trigonometric functions, implicit functions, exponential and logarithmic functions

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Implicit Differentiation is used when yy cannot be easily isolated as a function of xx. In equations like x2+y2=r2x^2 + y^2 = r^2, we differentiate every term with respect to xx, applying the chain rule to terms involving yy (treating yy as f(x)f(x)) to solve for dydx\frac{dy}{dx}.

A circle representing an implicit function where y is not uniquely defined for each x.
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The Exponential Function f(x)=exf(x) = e^x is unique because its derivative is equal to the function itself. Graphically, the slope of the tangent at any point (x,ex)(x, e^x) is exactly exe^x.

Graph of the exponential function e^x showing rapid growth.
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Logarithmic Differentiation is a technique used for functions of the form y=[f(x)]g(x)y = [f(x)]^{g(x)}. By taking the natural log (ln⁡)(\ln) of both sides, we use the property ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a to transform the power into a product, making it easier to differentiate.

Flowchart showing the steps of logarithmic differentiation.
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Inverse Trigonometric Functions have specific domains for differentiability. For instance, sin⁡−1x\sin^{-1} x is defined for x∈[−1,1]x \in [-1, 1], but its derivative exists only for x∈(−1,1)x \in (-1, 1) because the slope becomes infinite at the boundaries.

Graph of the natural logarithmic function.

📐Formulae

ddx(sin⁡−1x)=11−x2, for x∈(−1,1)\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}, \text{ for } x \in (-1, 1)

ddx(cos⁡−1x)=−11−x2, for x∈(−1,1)\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}, \text{ for } x \in (-1, 1)

ddx(tan⁡−1x)=11+x2, for x∈R\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}, \text{ for } x \in \mathbb{R}

ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

ddx(ax)=axln⁡a\frac{d}{dx}(a^x) = a^x \ln a

ddx(ln⁡x)=1x, for x>0\frac{d}{dx}(\ln x) = \frac{1}{x}, \text{ for } x > 0

ddx(log⁡ax)=1xln⁡a\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}

ddx(uv)=uv[vududx+ln⁡udvdx]\frac{d}{dx}(u^v) = u^v \left[ \frac{v}{u} \frac{du}{dx} + \ln u \frac{dv}{dx} \right]

💡Examples

Problem 1:

Find dydx\frac{dy}{dx} if x2+xy+y2=100x^2 + xy + y^2 = 100.

Solution:

  1. Differentiate both sides of the equation with respect to xx: ddx(x2)+ddx(xy)+ddx(y2)=ddx(100)\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(100)
  2. Apply the power rule to x2x^2 and the product rule to xyxy: 2x+(xdydx+y⋅1)+2ydydx=02x + (x \frac{dy}{dx} + y \cdot 1) + 2y \frac{dy}{dx} = 0
  3. Group the terms involving dydx\frac{dy}{dx}: xdydx+2ydydx=−2x−yx \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y
  4. Factor out dydx\frac{dy}{dx}: dydx(x+2y)=−(2x+y)\frac{dy}{dx}(x + 2y) = -(2x + y)
  5. Solve for dydx\frac{dy}{dx}: dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Explanation:

This is an implicit differentiation problem. We treat yy as a function of xx and use the Product Rule for the term xyxy and the Chain Rule for y2y^2.

Problem 2:

Differentiate y=xsin⁡xy = x^{\sin x} with respect to xx.

Solution:

  1. Since the variable is in both the base and the exponent, take the natural logarithm of both sides: ln⁡y=ln⁡(xsin⁡x)\ln y = \ln(x^{\sin x})
  2. Use the logarithm power property: ln⁡y=sin⁡x⋅ln⁡x\ln y = \sin x \cdot \ln x
  3. Differentiate both sides with respect to xx using the Product Rule on the right side: 1ydydx=ddx(sin⁡x)⋅ln⁡x+sin⁡x⋅ddx(ln⁡x)\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \ln x + \sin x \cdot \frac{d}{dx}(\ln x) 1ydydx=cos⁡xln⁡x+sin⁡x⋅1x\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x}
  4. Multiply by yy to isolate dydx\frac{dy}{dx}: dydx=y(cos⁡xln⁡x+sin⁡xx)\frac{dy}{dx} = y \left( \cos x \ln x + \frac{\sin x}{x} \right)
  5. Substitute the original expression for yy: dydx=xsin⁡x(cos⁡xln⁡x+sin⁡xx)\frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)

Explanation:

This problem requires logarithmic differentiation because the function is of the form u(x)v(x)u(x)^{v(x)}. Taking logs simplifies the exponent into a product.

Problem 3:

Find the derivative dydx\frac{dy}{dx} for the implicit function sin⁡y+x=cos⁡x\sin y + x = \cos x.

Mapping of terms to their derivatives.

Solution:

Differentiating both sides with respect to xx: ddx(sin⁡y)+ddx(x)=ddx(cos⁡x)\frac{d}{dx}(\sin y) + \frac{d}{dx}(x) = \frac{d}{dx}(\cos x) Using chain rule for sin⁡y\sin y: cos⁡y⋅dydx+1=−sin⁡x\cos y \cdot \frac{dy}{dx} + 1 = -\sin x Isolating dydx\frac{dy}{dx}: cos⁡y⋅dydx=−1−sin⁡x\cos y \cdot \frac{dy}{dx} = -1 - \sin x dydx=−1+sin⁡xcos⁡y\frac{dy}{dx} = -\frac{1 + \sin x}{\cos y}

Explanation:

We apply the chain rule to the term involving yy and then use algebraic manipulation to solve for the derivative term.

Problem 4:

Differentiate y=ecos⁡xy = e^{\cos x} with respect to xx.

Graph showing the periodic nature of e^cos(x).

Solution:

Let u=cos⁡xu = \cos x. Then y=euy = e^u. Using the chain rule: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} dydu=eu=ecos⁡x\frac{dy}{du} = e^u = e^{\cos x} dudx=−sin⁡x\frac{du}{dx} = -\sin x Therefore, dydx=ecos⁡x⋅(−sin⁡x)=−sin⁡x⋅ecos⁡x\frac{dy}{dx} = e^{\cos x} \cdot (-\sin x) = -\sin x \cdot e^{\cos x}

Explanation:

This demonstrates the chain rule applied to an exponential function where the exponent is a trigonometric function.