Differentiation – JEE Main MathematicsPractice Questions & PYQs

To determine the derivative of $f(f(f(x)))+(f(x))^{2}$ at $x=1$, we apply the chain rule separately to each term. The derivative of the triple composition $f(f(f(x)))$ is calculated as $f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$, while the derivative of the power term $(f(x))^{2}$ follows the power rule as $2f(x) \cdot f'(x)$. Evaluating these expressions at $x=1$, where $f(1)=1$, simplifies the composite chain to $f'(1) \cdot f'(1) \cdot f'(1)$ and the power term to $2f(1) \cdot f'(1)$. Using the provided values $f(1)=1$ and $f'(1)=3$, the calculation becomes $3 \cdot 3 \cdot 3 + 2 \cdot 1 \cdot 3$, which results in $27+6$ for a final value of 33.

Dividing both the numerator and the denominator of the internal fraction by 2 reveals standard trigonometric ratios, where $\frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right)$ and $\frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$. Substituting these values transforms the expression into $\frac{\sin\left(\frac{\pi}{3}\right)\cos x + \cos\left(\frac{\pi}{3}\right)\sin x}{\cos\left(\frac{\pi}{3}\right)\cos x - \sin\left(\frac{\pi}{3}\right)\sin x}$, which simplifies using angle sum formulas to $\frac{\sin\left(\frac{\pi}{3} + x\right)}{\cos\left(\frac{\pi}{3} + x\right)}$, or $\tan\left(\frac{\pi}{3} + x\right)$. Given that $\cot^{-1}(\tan \theta) = \frac{\pi}{2} - \theta$, the original relationship becomes $2y = \left( \frac{\pi}{2} - \left( \frac{\pi}{3} + x \right) \right)^2$, which reduces to $2y = \left( \frac{\pi}{6} - x \right)^2$. Differentiating this expression with respect to $x$ by applying the chain rule produces $2 \frac{dy}{dx} = 2 \left( \frac{\pi}{6} - x \right) \cdot (-1)$, which resolves to the final derivative of $\frac{dy}{dx} = x - \frac{\pi}{6}$.

Simplifying the expression involves dividing the numerator and denominator by $\cos x$ to express the argument of the inverse tangent as $\frac{\tan x - 1}{\tan x + 1}$, which is identical to $\tan(x - \frac{\pi}{4})$. Because the inverse tangent function cancels with the tangent function, the original expression reduces simply to $x - \frac{\pi}{4}$, and its derivative with respect to $x$ is 1. The requirement is to differentiate this with respect to $\frac{x}{2}$ rather than $x$, which involves dividing the derivative of the original expression by the derivative of $\frac{x}{2}$. Since the derivative of $\frac{x}{2}$ is $\frac{1}{2}$, the final result is:

$\frac{1}{\frac{1}{2}} = 2$

Implicit differentiation is the standard method for determining derivatives of equations where variables remain intertwined rather than explicitly separated. Differentiating $e^{y} + xy = e$ with respect to $x$ gives $e^{y} \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$, and setting $x=0$ leads to $e^{y}=e$, or $y=1$, which simplifies the first derivative evaluation to $e \frac{dy}{dx} + 1 = 0$, yielding $\frac{dy}{dx} = -\frac{1}{e}$. Differentiating the equation a second time requires applying the product rule to each term to obtain $e^{y} \left(\frac{dy}{dx}\right)^{2} + e^{y} \frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + x \frac{d^{2}y}{dx^{2}} = 0$. Substituting the values $x=0, y=1,$ and $\frac{dy}{dx} = -\frac{1}{e}$ into this expression results in $\frac{1}{e} + e \frac{d^{2}y}{dx^{2}} - \frac{2}{e} = 0$, establishing that $e \frac{d^{2}y}{dx^{2}} = \frac{1}{e}$, or $\frac{d^{2}y}{dx^{2}} = \frac{1}{e^{2}}$.

Using the inverse trigonometric identity $\text{cosec}^{-1}t + \text{sec}^{-1}t = \frac{\pi}{2}$ for $|t| \geq 1$, the product of the squared expressions $x^2 = 2^{\text{cosec}^{-1}t}$ and $y^2 = 2^{\text{sec}^{-1}t}$ simplifies to $x^2 y^2 = 2^{\text{cosec}^{-1}t + \text{sec}^{-1}t} = 2^{\frac{\pi}{2}}$. Since $2^{\frac{\pi}{2}}$ is a constant, differentiating the implicit relation $x^2 y^2 = 2^{\frac{\pi}{2}}$ with respect to $x$ using the product rule gives $2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$. Solving for the derivative yields $2x^2 y \frac{dy}{dx} = -2x y^2$, which simplifies to $\frac{dy}{dx} = -\frac{y}{x}$.

L'Hôpital's Rule states that if $f(0)=0$ and $f'(0)=0$, the limit $\lim_{x\to 0} \frac{f'(x)}{x}$ is equal to $f''(0)$. The derivative of the determinant $f(x)$ is the sum of determinants obtained by differentiating each row sequentially, and evaluating this at $x=0$ shows that $f'(0)=0$ because of row dependence and zero-value rows.

Evaluation of the second derivative $f''(0)$ utilizes the formula

. Using the values $R_1(0)=(1,0,1)$, $R_2(0)=(0,0,0)$, $R_3(0)=(0,0,1)$, $R_1'(0)=(0,1,0)$, $R_2'(0)=(2,0,2)$, $R_3'(0)=(1,1,0)$, $R_1''(0)=(-1,0,0)$, $R_2''(0)=(0,2,0)$, and $R_3''(0)=(0,0,0)$, the expression simplifies to:
$f''(0) = 0 + \begin{vmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{vmatrix} + 0 + 2\begin{vmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 0 & 1 \end{vmatrix} + 0 + 2\begin{vmatrix} 1 & 0 & 1 \\ 2 & 0 & 2 \\ 1 & 1 & 0 \end{vmatrix}$
Simplification of the non-zero determinants yields $2 + 2(-2) + 0$, which equals $-2$.

Implicit differentiation of the equation $x^2 + y^2 + \sin y = 4$ with respect to $x$ yields $2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$. Substituting the point $(-2,0)$ into this expression results in $2(-2) + 2(0) \frac{dy}{dx} + \cos(0) \frac{dy}{dx} = 0$, which simplifies to $-4 + \frac{dy}{dx} = 0$, identifying the first derivative value as $\frac{dy}{dx} = 4$. Differentiating the implicit expression $2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$ again with respect to $x$ produces $2 + 2 \left( \frac{dy}{dx} \right)^2 + 2y \frac{d^2y}{dx^2} - \sin y \left( \frac{dy}{dx} \right)^2 + \cos y \frac{d^2y}{dx^2} = 0$. Substituting $x = -2$, $y = 0$, and $\frac{dy}{dx} = 4$ into this second-order equation gives $2 + 2(4)^2 + 2(0) \frac{d^2y}{dx^2} - \sin(0)(4)^2 + \cos(0) \frac{d^2y}{dx^2} = 0$. This reduces to $2 + 32 + \frac{d^2y}{dx^2} = 0$, which results in $\frac{d^2y}{dx^2} = -34$.

Assuming $f(x) = ax^n$ to satisfy the polynomial degree requirement, the derivatives are $f'(x) = nax^{n-1}$ and $f''(x) = n(n-1)ax^{n-2}$. Substituting these into the expression $f(3x) = f'(x) \cdot f''(x)$ gives $a(3x)^n = (nax^{n-1})(n(n-1)ax^{n-2})$, which simplifies to $a 3^n x^n = n^2(n-1)a^2 x^{2n-3}$. Equating the powers of $x$ gives $n = 2n - 3$, or $n = 3$, and equating coefficients yields $27a = 3^2(3-1)a^2$, which simplifies to $27a = 18a^2$, resulting in $a = \frac{3}{2}$.

With the function defined as $f(x) = \frac{3}{2}x^3$, its derivatives are $f'(x) = \frac{9}{2}x^2$ and $f''(x) = 9x$. Evaluating these at $x=2$ results in $f(2) = \frac{3}{2}(8) = 12$, $f'(2) = \frac{9}{2}(4) = 18$, and $f''(2) = 9(2) = 18$. Substituting these values into the expression $f''(2) - f'(2)$ yields $18 - 18 = 0$.

Defining $u = x+\sqrt{x^{2-}1}$ and $v = x-\sqrt{x^{2-}1}$ allows the expression $y = u^{15} + v^{15}$ to be differentiated with respect to $x$ using the chain rule. Because $u \cdot v = 1$, we can express $v$ as $u^{-1}$, which simplifies $y$ to $u^{15} + u^{-15}$. Calculating the first derivative yields $\frac{dy}{dx} = 15(u^{14} - u^{-14}) \frac{du}{dx}$, where $\frac{du}{dx} = \frac{u}{\sqrt{x^{2-}1}}$. Substituting this derivative back into the expression simplifies the first derivative to $\frac{dy}{dx} = \frac{15(u^{15}-u^{-15})}{\sqrt{x^{2-}1}}$.

Setting $z = u^{15} - u^{-15}$, the relationship becomes $\sqrt{x^{2-}1} \frac{dy}{dx} = 15z$. Differentiating both sides of this equation with respect to $x$ gives $\frac{x}{\sqrt{x^{2-}1}} \frac{dy}{dx} + \sqrt{x^{2-}1} \frac{d^2y}{dx^2} = 15 \frac{dz}{dx}$. Since $\frac{dz}{dx} = 15(u^{14} + u^{-14}) \frac{du}{dx} = \frac{15(u^{15}+u^{-15})}{\sqrt{x^{2-}1}}$, which is simply $\frac{15y}{\sqrt{x^{2-}1}}$, the equation becomes $\frac{x}{\sqrt{x^{2-}1}} \frac{dy}{dx} + \sqrt{x^{2-}1} \frac{d^2y}{dx^2} = \frac{225y}{\sqrt{x^{2-}1}}$. Multiplying the entire equation by $\sqrt{x^{2-}1}$ confirms that $(x^{2-}1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 225y$.

The core concept here relies on recognizing that the argument inside the inverse tangent function follows the pattern of the double-angle identity for tangent, specifically $\tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right) = 2\tan^{-1}(\theta)$. By setting $\theta = 3x^{3/2}$, the original expression becomes $\tan^{-1}\left(\frac{2(3x^{3/2})}{1-(3x^{3/2})^2}\right)$, which simplifies directly to $2\tan^{-1}(3x^{3/2})$.

Differentiating this simplified form requires applying the chain rule, where the derivative of $2\tan^{-1}(3x^{3/2})$ is $2 \cdot \frac{1}{1+(3x^{3/2})^2}$ multiplied by the derivative of the inner term $3x^{3/2}$. Calculating the derivative of $3x^{3/2}$ yields $3 \cdot \frac{3}{2} \cdot x^{1/2}$, or $\frac{9}{2}\sqrt{x}$. Combining these components results in $\frac{2}{1+9x^3} \cdot \frac{9}{2}\sqrt{x}$, which simplifies to $\frac{9\sqrt{x}}{1+9x^3}$. Factoring out the $\sqrt{x}$ as required by the problem statement leaves $g(x)$ equal to $\frac{9}{1+9x^3}$.

The core principle linking a function $f$ and its inverse $g$ is the identity $f(g(x)) = x$. Differentiating both sides of this equation with respect to $x$ using the chain rule yields $f'(g(x)) \cdot g'(x) = 1$, which implies that the derivative of the inverse is simply the reciprocal of the function's derivative evaluated at $g(x)$. Substituting the given $f'(x) = \frac{1}{1+x^5}$ into this relation gives $f'(g(x)) = \frac{1}{1+(g(x))^5}$. Dividing one by this expression results in $g'(x) = 1+(g(x))^5$.

Differentiating the composite function $y=\sec(\tan^{-1}x)$ requires applying the chain rule, which involves taking the derivative of the outer secant function multiplied by the derivative of the inner inverse tangent function. Since the derivative of $\sec(u)$ is $\sec(u)\tan(u)$ and the derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, the complete derivative is expressed as $\frac{dy}{dx}=\sec(\tan^{-1}x)\tan(\tan^{-1}x) \cdot \frac{1}{1+x^2}$.

Evaluating this expression at $x=1$ involves substituting the value directly, where the inverse tangent component $\tan^{-1}(1)$ corresponds to the angle $\frac{\pi}{4}$. At this angle, the secant value $\sec(\frac{\pi}{4})$ is $\sqrt{2}$ and the tangent value $\tan(\frac{\pi}{4})$ is 1, while the rational component $\frac{1}{1+x^2}$ simplifies to $\frac{1}{2}$. Combining these results gives $\sqrt{2} \times 1 \times \frac{1}{2}$, which simplifies to $\frac{1}{\sqrt{2}}$.

Understanding the second derivative of an inverse function relies on applying the chain rule to the relationship $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$ while differentiating with respect to $y$. Because the variable of differentiation is $y$, we express the derivative operator $\frac{d}{dy}$ as the product $\frac{dx}{dy} \frac{d}{dx}$, which allows us to differentiate the reciprocal expression with respect to $x$.

When performing this operation, the chain rule yields $-\left(\frac{dy}{dx}\right)^{-2} \frac{d^2 y}{dx^2}$ by applying the power rule to the first derivative. Multiplying this result by the remaining factor of $\frac{dx}{dy}$, which is equivalent to $\left(\frac{dy}{dx}\right)^{-1}$, simplifies the entire expression to $- \left(\frac{d^2 y}{dx^2}\right) \left(\frac{dy}{dx}\right)^{-3}$.

The chain rule is the essential tool for differentiating a composite function, requiring us to peel back each layer starting from the outermost component. For the expression $g(x)=[f(2f(x)+2)]^2$, the derivative $g'(x)$ begins by applying the power rule to the outer square, resulting in $2f(2f(x)+2)$ multiplied by the derivative of the inner function. This inner derivative is found by applying the chain rule again, which produces $f'(2f(x)+2)$ multiplied by the derivative of the nested term $2f(x)+2$, specifically $2f'(x)$. Combining these components yields the full derivative:

$g'(x) = 4 f(2f(x)+2) \cdot f'(2f(x)+2) \cdot f'(x)$

Evaluating this at $x=0$ requires substituting the known values $f(0)=-1$ and $f'(0)=1$. The argument within the function simplifies to $2f(0)+2 = 2(-1)+2 = 0$, allowing for a direct numerical evaluation. Substituting these values into the derivative formula provides the final result:

$g'(0) = 4 f(0) \cdot f'(0) \cdot f'(0) = 4(-1)(1)^2 = -4$

Implicit differentiation becomes significantly more manageable if we first isolate the trigonometric term, rewriting the equation $x^{2x} - 2x^x \cot y - 1 = 0$ as $2\cot y = x^x - x^{-x}$. Differentiating both sides with respect to $x$ requires the chain rule on the left side and the derivative of powers of $x$ on the right, producing the following relationship:

$-2\csc^2 y \cdot y' = (x^x + x^{-x})(1 + \log x)$

Substituting $x=1$ into the original rearranged equation gives $2\cot y = 1^1 - 1^{-1} = 0$, which establishes that $\cot y = 0$. Because $\csc^2 y = 1 + \cot^2 y$, we find that $\csc^2 y = 1 + 0^2 = 1$ at this point. Substituting these values into the derivative expression yields $-2(1) \cdot y' = (1 + 1)(1 + \log 1)$, which simplifies to $-2y' = 2$. Consequently, $y'(1) = -1$.

Applying natural logarithms to both sides of the equation simplifies the exponentiation and product, transforming the expression into $m \ln x + n \ln y = (m+n) \ln (x+y)$. This strategy makes differentiation significantly more manageable by converting the implicit relationship into a linear form where the power rule is no longer required for the outer functions.

Differentiating both sides with respect to $x$ provides the equation $\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left(1+ \frac{dy}{dx}\right)$. Expanding the right side and grouping the terms involving $\frac{dy}{dx}$ to one side allows for simple rearrangement into $\frac{dy}{dx} \left( \frac{n}{y} - \frac{m+n}{x+y} \right) = \frac{m+n}{x+y} - \frac{m}{x}$. After finding a common denominator for both sides and simplifying the algebraic expressions, the common factors \in the numerators reduce, eventually yielding the result $\frac{dy}{dx} = \frac{y}{x}$.

For any second-degree polynomial defined as $f(x) = Ax^2 + Bx + C$, the condition $f(1) = f(-1)$ implies $A(1)^2 + B(1) + C = A(-1)^2 + B(-1) + C$. Simplifying this equation gives $A + B + C = A - B + C$, which confirms that the linear coefficient $B$ must be zero, reducing the function to $f(x) = Ax^2 + C$ and its derivative to $f'(x) = 2Ax$. Evaluating this derivative at three terms $a$, $b$, and $c$ that form an Arithmetic Progression results in the values $2Aa$, $2Ab$, and $2Ac$. Because $a$, $b$, and $c$ share a common difference, their scalar multiples by $2A$ necessarily preserve that same proportional relationship, confirming that the resulting derivative values form an Arithmetic Progression.

The pattern in this series corresponds directly to the expansion of a binomial power. Computing the $k$-th derivative of $f(x)=x^n$ at $x=1$ yields $f^{(k)}(1) = \frac{n!}{(n-k)!}$, and dividing this value by $k!$ isolates the binomial coefficient, $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Replacing every term in the given series with its equivalent binomial coefficient creates the alternating sum $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n}$. By the binomial theorem, this alternating series represents the expansion of $(1-1)^n$. Since $(1-1)^n = 0^n$, the value of the entire expression is zero.

Finding the value of $(1+x^2)\frac{d^2 y}{dx^2}+x\frac{dy}{dx}$ involves establishing a differential relationship by differentiating the function and eliminating the square root. Start by calculating the first derivative of $y=(x+\sqrt{1+x^2})^n$ using the chain rule, which results in $\frac{dy}{dx} = n(x+\sqrt{1+x^2})^{n-1} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}}\right)$. Upon simplification, this becomes $\frac{dy}{dx} = \frac{ny}{\sqrt{1+x^2}}$, or alternatively $\sqrt{1+x^2} \frac{dy}{dx} = ny$. Squaring both sides yields $(1+x^2) \left(\frac{dy}{dx}\right)^2 = n^2 y^2$. Differentiating this result with respect to $x$ leads to $(1+x^2) \cdot 2 \frac{dy}{dx} \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \cdot 2x = n^2 \cdot 2y \frac{dy}{dx}$. Dividing the entire equation by $2 \frac{dy}{dx}$ reduces the expression to $(1+x^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = n^2 y$.