Derivatives | CBSE - Wyatt's Notes

sources:

• text: Standard textbook reference

Derivatives measure the rate of change of a function with respect to its variable. This topic covers basic rules, chain rule, implicit differentiation, and applications.

• $\frac{d}{dx}[x^n] = nx^{n-1}$ (power rule)
• $\frac{d}{dx}[e^x] = e^x$, $\frac{d}{dx}[\ln x] = \frac{1}{x}$
• $\frac{d}{dx}[\sin x] = \cos x$, $\frac{d}{dx}[\cos x] = -\sin x$
• $\frac{d}{dx}[\tan x] = \sec^2 x$
• $\frac{d}{dx}[\csc x] = -\csc x \cot x$, $\frac{d}{dx}[\sec x] = \sec x \tan x$
• $\frac{d}{dx}[\cot x] = -\csc^2 x$
• Product rule: $\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$
• Quotient rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
• Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Problem: Find $\frac{d}{dx}[x^2 \sin x]$.

Solution:

Let $u = x^2$ and $v = \sin x$. Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \cos x$.

$\frac{d}{dx}[x^2 \sin x] = x^2 \cos x + \sin x \cdot 2x = x^2 \cos x + 2x \sin x$

Common mistake: Forgetting to differentiate one of the factors. Both terms must be present.

Problem: Find $\frac{d}{dx}[\sin(3x^2 + 1)]$.

Solution:

Let $u = 3x^2 + 1$, so $\frac{du}{dx} = 6x$.

$\frac{d}{dx}[\sin(3x^2 + 1)] = \cos(3x^2 + 1) \cdot 6x = 6x \cos(3x^2 + 1)$

Common mistake: Forgetting the inner derivative. The answer is not just $\cos(3x^2 + 1)$.

Problem: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$.

Solution:

Differentiate both sides with respect to $x$: $2x + 2y\frac{dy}{dx} = 0$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x}{y}$

Common mistake: Treating $y$ as a constant when differentiating. Remember that $y$ is a function of $x$, so $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$.

Problem: Find $\frac{d}{dx}\left[\frac{x}{x^2 + 1}\right]$.

Solution:

Let $u = x$ and $v = x^2 + 1$. Then $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = 2x$.

$\frac{d}{dx}\left[\frac{x}{x^2+1}\right] = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$

Common mistake: Swapping $u$ and $v$ in the quotient rule formula. The numerator is $v \cdot u' - u \cdot v'$, not $u \cdot v' - v \cdot u'$.

Problem: If $y = e^{2x}$, find $\frac{d^2y}{dx^2}$.

Solution:

First derivative: $\frac{dy}{dx} = 2e^{2x}$

Second derivative: $\frac{d^2y}{dx^2} = 2 \cdot 2e^{2x} = 4e^{2x}$

Common mistake: Forgetting to apply the chain rule at each differentiation step. Each derivative of $e^{2x}$ brings down a factor of 2.

1. Find $\frac{d}{dx}[x^3 e^x]$ using the product rule.
2. Find $\frac{d}{dx}[\ln(\sin x)]$ using the chain rule.
3. Find $\frac{dy}{dx}$ if $xy + y^2 = 10$ using implicit differentiation.
4. Find $\frac{d}{dx}\left[\frac{\sin x}{1 + \cos x}\right]$ using the quotient rule.
5. Find the second derivative of $f(x) = x \ln x$.

• Identify which rule to apply before differentiating
• For chain rule, always differentiate the outer function first, then multiply by derivative of inner function
• Practice implicit differentiation with equations of circles and ellipses
• Check answers by substituting back into the original equation when possible
• Higher-order derivatives require applying the differentiation rule repeatedly
• When a function is a product of more than two factors, differentiate one pair at a time

1. Write down $u$ and $v$ (or $u$ and $v$ for quotient rule) explicitly before computing derivatives.
2. For chain rule, identify the outermost function and work inward layer by layer.
3. In implicit differentiation, collect all $\frac{dy}{dx}$ terms on one side and factor.
4. Simplify your answer by factoring common terms; examiners reward simplified forms.
5. Verify your derivative by checking dimensions or testing a specific value.

Problem: Find $\frac{d}{dx}[x \cdot e^x \cdot \sin x]$.

Solution:

Apply the product rule iteratively. Let $u = x$, $v = e^x \sin x$.

First, find $\frac{dv}{dx}$ using the product rule on $e^x \sin x$: $\frac{d}{dx}[e^x \sin x] = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$

Now apply the product rule to $x \cdot (e^x \sin x)$: $\frac{d}{dx}[x \cdot e^x \sin x] = 1 \cdot e^x \sin x + x \cdot e^x(\sin x + \cos x)$

Factor: $= e^x[\sin x + x\sin x + x\cos x] = e^x[(1 + x)\sin x + x\cos x]$

Common mistake: Trying to apply the product rule to three factors at once. Instead, group two factors together and apply the rule iteratively.

Problem: Find $\frac{dy}{dx}$ if $y = x^x$ for $x > 0$.

Solution:

Take the natural logarithm of both sides: $\ln y = \ln(x^x) = x \ln x$

Differentiate both sides with respect to $x$: $\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Common mistake: Treating $x^x$ as a power function ($nx^{n-1}$) or an exponential function ($a^x \ln a$). It is neither — use logarithmic differentiation.

Problem: If $f(x) = \tan^{-1}\left(\frac{x}{1 + x^2}\right)$, find $f'(x)$.

Solution:

Let $u = \frac{x}{1 + x^2}$. Then $f(x) = \tan^{-1}(u)$ and: $f'(x) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$

Compute $\frac{du}{dx}$ using the quotient rule: $\frac{du}{dx} = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2}$

Now compute $\frac{1}{1 + u^2}$: $1 + u^2 = 1 + \frac{x^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 + x^2}{(1 + x^2)^2} = \frac{1 + 2x^2 + x^4 + x^2}{(1 + x^2)^2} = \frac{1 + 3x^2 + x^4}{(1 + x^2)^2}$

Therefore: $f'(x) = \frac{(1 + x^2)^2}{1 + 3x^2 + x^4} \cdot \frac{1 - x^2}{(1 + x^2)^2} = \frac{1 - x^2}{1 + 3x^2 + x^4}$

Common mistake: Forgetting the chain rule when differentiating inverse trigonometric functions. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^2} \cdot \frac{du}{dx}$, not just $\frac{1}{1 + u^2}$.

1. For logarithmic differentiation, take $\ln$ of both sides first, then differentiate implicitly. This is useful for functions of the form $f(x)^{g(x)}$.
2. The derivative of $\ln|f(x)|$ is $\frac{f'(x)}{f(x)}$ — remember the absolute value for the domain.
3. When differentiating inverse trigonometric functions, always apply the chain rule: $\frac{d}{dx}[\tan^{-1}(u)] = \frac{u'}{1 + u^2}$.
4. For piecewise functions, check differentiability at the boundary by verifying that the left and right derivatives are equal.
5. In CBSE exams, questions often combine multiple rules (product + chain, quotient + chain). Practice identifying which rules to apply in which order.

A derivative measures how fast something changes at a single instant. Think of it as the slope of a tangent line touching a curve — it tells you whether the function is going up, going down, or flat at that point. The power rule is the simplest case: the derivative of x^n is n times x^(n-1), which follows from the pattern of how polynomial rates work. The chain rule handles nested functions: if something depends on something else that depends on x, you multiply their individual rates of change. Implicit differentiation is useful when y is buried inside an equation and you cannot isolate it efficiently.

• Integrals — the reverse of differentiation
• Matrices — linear algebra foundations
• CBSE Physics — applications of derivatives in mechanics

Confusing the product rule and chain rule. The product rule (uv)’ = u’v + uv’ applies to the product of two functions. The chain rule (f(g(x)))’ = f’(g(x)) * g’(x) applies to composition. Students often apply the product rule to composite functions or the chain rule to products, giving incorrect derivatives.

Forgetting the negative sign in trigonometric derivatives. The derivatives of cos(x), cot(x), csc(x), and their reciprocals all have negative signs. Students frequently forget these negatives, especially for cos(x) -> -sin(x) and csc(x) -> -csc(x)cot(x). This is the most common source of sign errors in differentiation.

Incorrectly applying the chain rule to nested functions. When differentiating sin(x^2), the result is cos(x^2) * 2x, not just cos(x^2). Students often differentiate the outer function but forget to multiply by the derivative of the inner function. The chain rule must be applied at every level of nesting.