krit.club logo

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

•

Implicit Differentiation: This technique is used when yy cannot be easily expressed as an explicit function of xx (e.g., x2+y2=r2x^2 + y^2 = r^2). We differentiate every term with respect to xx, applying the Chain Rule to terms involving yy by multiplying by dydx\frac{dy}{dx}. Visually, this allows us to find the slope of the tangent at any point (x,y)(x, y) on a curve, even if the curve fails the vertical line test (like a circle).

•

Derivatives of Inverse Trigonometric Functions: These derivatives are defined within specific restricted domains to ensure the functions are one-to-one. For example, the derivative of sin⁡−1x\sin^{-1} x is 11−x2\frac{1}{\sqrt{1-x^2}}. Visually, the graph of sin⁡−1x\sin^{-1} x is an increasing curve that becomes steeper as xx approaches 11 or −1-1, where the slope becomes infinite, explaining why the derivative is undefined at those boundaries.

•

Exponential Functions: The derivative of the natural exponential function exe^x is unique because it is its own derivative, ddx(ex)=ex\frac{d}{dx}(e^x) = e^x. Graphically, this means the height of the curve at any point is exactly equal to the slope of the tangent line at that point. For any other base a>0a > 0, the derivative is axln⁡aa^x \ln a, where the ln⁡a\ln a acts as a vertical scaling factor for the slope.

•

Logarithmic Functions: The derivative of ln⁡x\ln x is 1x\frac{1}{x} for x>0x > 0. Visually, as xx increases, the slope of the logarithmic curve 1x\frac{1}{x} decreases, indicating that the function grows slower and slower. The function is only defined for positive xx, and its derivative shows that the curve is always increasing but concave down.

•

Logarithmic Differentiation: This method is essential for functions where the variable is in both the base and the exponent, such as y=[f(x)]g(x)y = [f(x)]^{g(x)}, or for products of many terms. By taking the natural log of both sides, we use the property ln⁡(uv)=vln⁡u\ln(u^v) = v \ln u to transform an exponential relationship into a product, making it easier to differentiate using the Product Rule and Chain Rule.

•

Chain Rule for Composite Functions: When dealing with nested functions like ln⁡(sin⁡x)\ln(\sin x), we differentiate from the outside in. Visually, this is like examining how a change in the input xx ripples through a sequence of transformations. If y=f(u)y = f(u) and u=g(x)u = g(x), then the total rate of change dydx\frac{dy}{dx} is the product of the rate of change of each 'layer': dydu⋅dudx\frac{dy}{du} \cdot \frac{du}{dx}.

📐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.