Differentiation – JEE Main MathematicsPractice Questions & PYQs

Determining the second derivative of a function involving products requires consistent application of the product rule and the chain rule. Starting with the first derivative $f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right)$, we differentiate the components to obtain the full expression for the second derivative:
$f''(x) = 6x \sin\left(\frac{1}{x}\right) - 4 \cos\left(\frac{1}{x}\right) - \frac{1}{x} \sin\left(\frac{1}{x}\right)$
Substituting $x = \frac{2}{\pi}$ into this expression, where $\sin\left(\frac{\pi}{2}\right) = 1$ and $\cos\left(\frac{\pi}{2}\right) = 0$, the calculation results in $6\left(\frac{2}{\pi}\right)(1) - 4(0) - \left(\frac{\pi}{2}\right)(1)$. This simplifies to the difference $\frac{12}{\pi} - \frac{\pi}{2}$, which equates to
$f''\left(\frac{2}{\pi}\right) = \frac{24-\pi^2}{2\pi}$

Since $f(x)=x^{2+}g'(1)x+g''(2)$, we identify the derivatives as $f'(x)=2x+g'(1)$ and $f''(x)=2$. Substituting these expressions into the definition of $g(x)$ yields $g(x)=f(1)x^{2+}x(2x+g'(1))+2$, which simplifies to $g(x)=(f(1)+2)x^{2+}g'(1)x+2$. Because $g(x)$ is a quadratic polynomial, its derivative is $g'(x)=2(f(1)+2)x+g'(1)$. Setting $x=1$ in this derivative allows us to equate $g'(1)=2(f(1)+2)(1)+g'(1)$, which simplifies directly to $2(f(1)+2)=0$, forcing $f(1)=-2$.

Given $f(1)=-2$, the second derivative $g''(x)=2(f(1)+2)$ is zero, which means $g''(2)=0$. Returning to the definition $f(1)=1^{2+}g'(1)(1)+g''(2)$, we substitute the known values to get $-2=1+g'(1)+0$, resulting in $g'(1)=-3$. With these coefficients determined, the functions are $f(x)=x^{2-}3x$ and $g(x)=-3x+2$. Evaluating both at $x=4$, we find $f(4)=4^{2-}3(4)=4$ and $g(4)=-3(4)+2=-10$. The difference $f(4)-g(4)$ is therefore $4-(-10)$, which equals 14.

Solving functional equations involving both $f(x)$ and $f(1/x)$ is most effectively handled by treating the expressions as a system of linear equations. By substituting $x$ with $1/x$ in the original equation, you generate a second independent relation, which allows for the algebraic elimination of $f(1/x)$ to isolate $f(x)$.

The original equation is given by:
$3 f(x) + 2 f \left( \frac{1}{x} \right) = \frac{1}{x} - 10$
Substituting $1/x$ for $x$ transforms the equation into $3 f(1/x) + 2 f(x) = x - 10$. Multiplying the first equation by 3 and the second by 2 produces $9 f(x) + 6 f(1/x) = 3/x - 30$ and $4 f(x) + 6 f(1/x) = 2x - 20$. Subtracting the second from the first yields $5 f(x) = 3/x - 2x - 10$, which simplifies to $f(x) = \frac{1}{5} \left( \frac{3}{x} - 2x - 10 \right)$.

Differentiating this expression with respect to $x$ results in $f'(x) = \frac{1}{5} \left( -\frac{3}{x^2} - 2 \right)$. Evaluating the function at the specific points gives $f(3) = \frac{1}{5} (1 - 6 - 10) = -3$, while calculating the derivative at $1/4$ yields $f'(1/4) = \frac{1}{5} (-48 - 2) = -10$. Taking the absolute value of their sum, $|-3 - 10|$, results in a final value of 13.

The product $y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$ follows a telescoping pattern that simplifies elegantly when multiplied by the conjugate $(1-x)$. This operation yields the identity $(1-x)y = 1-x^{32}$, or $y - xy = 1 - x^{32}$, which allows us to differentiate the function without expanding the entire polynomial. Differentiating both sides with respect to $x$ using the product rule leads to the first-order relation $y' - y - xy' = -32x^{31}$.

Differentiating this result once more gives $y'' - y' - y' - xy'' = -32(31)x^{30}$, which simplifies to $y'' - 2y' - xy'' = -992x^{30}$. By evaluating these derivatives at $x = -1$, the first equation becomes $2y' - y = 32$, and since $y$ at $x = -1$ is $0$, we find that $y' = 16$. Substituting these values into the second-order equation $2y'' - 2y' = -992$ and simplifying confirms the final result for $y' - y''$ is $496$.

When a function involves a variable in both the base and the exponent, logarithmic differentiation is the most effective approach because it enables the handling of terms like $x^y$ by treating them as $e^{y \ln x}$. Applying the derivative operator $\frac{d}{dx}$ to the equation $2 x^y + 3 y^x = 20$ results in $2 x^y \left( \frac{y}{x} + \ln x \frac{dy}{dx} \right) + 3 y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) = 0$.

Substituting the coordinates $(2,2)$ into this equation simplifies the terms significantly, as $2^2 = 4$ for both $x^y$ and $y^x$. This yields $8 \left( 1 + \ln 2 \frac{dy}{dx} \right) + 12 \left( \ln 2 + \frac{dy}{dx} \right) = 0$. Distributing the coefficients provides $8 + 8 \ln 2 \frac{dy}{dx} + 12 \ln 2 + 12 \frac{dy}{dx} = 0$. Grouping the terms containing $\frac{dy}{dx}$ leads to $\frac{dy}{dx} (12 + 8 \ln 2) = -(8 + 12 \ln 2)$. Factoring 4 from both the numerator and the denominator results in $\frac{dy}{dx} = - \frac{2 + 3 \ln 2}{3 + 2 \ln 2}$, which simplifies using logarithmic power rules to $- \left( \frac{2 + \ln 8}{3 + \ln 4} \right)$.

This problem requires applying the chain rule sequentially to navigate the composite function structure. By defining $u = \frac{\pi}{3}\cos(g(x))$ and $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^{3+}5x^{2+}1)^{\frac{3}{2}}$, the derivative of the function $y = \sin^3(u)$ is expressed as $y' = 3\sin^2(u)\cos(u) \cdot u'$, where $u' = -\frac{\pi}{3}\sin(g(x)) \cdot g'(x)$.

At $x=1$, we calculate $g(1) = \frac{\pi}{3\sqrt{2}}(2)^{\frac{3}{2}} = \frac{2\pi}{3}$. Differentiating the inner function yields $g'(x) = \frac{\pi}{2\sqrt{2}}(-4x^{3+}5x^{2+}1)^{\frac{1}{2}}(-12x^{2+}10x)$, which simplifies at $x=1$ to $g'(1) = \frac{\pi}{2\sqrt{2}}(\sqrt{2})(-2) = -\pi$. Using these values, we find $u(1) = \frac{\pi}{3}\cos(\frac{2\pi}{3}) = -\frac{\pi}{6}$ and consequently $y(1) = \sin^3(-\frac{\pi}{6}) = -1/8$.

To determine $y'(1)$, we evaluate $u'(1) = -\frac{\pi}{3}\sin(\frac{2\pi}{3})(-\pi) = \frac{\pi^2\sqrt{3}}{6}$. Substituting these components into the derivative formula gives $y'(1) = 3\sin^2(-\frac{\pi}{6})\cos(-\frac{\pi}{6}) \cdot \frac{\pi^2\sqrt{3}}{6} = 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\pi^2\sqrt{3}}{6} = \frac{3\pi^2}{16}$. Combining these results into the linear expression $2y'(1) + 3\pi^2 y(1)$ leads to $2(\frac{3\pi^2}{16}) + 3\pi^2(-1/8) = \frac{3\pi^2}{8} - \frac{3\pi^2}{8} = 0$.

Defining the relationship between two twice differentiable functions by examining their difference allows us to evaluate properties of the system through successive integration. Given the second derivative relationship $f''(x) - g''(x) = 6x$, we integrate once to obtain $f'(x) - g'(x) = 3x^2 + C$. Using the initial conditions $f'(1) = 9$ and $4g'(1) - 3 = 9$, we determine $g'(1) = 3$, which leads to $f'(1) - g'(1) = 6$. Substituting $x = 1$ into our first derivative expression gives $3(1)^2 + C = 6$, yielding $C = 3$. Integrating once more provides $f(x) - g(x) = x^3 + 3x + D$. With $f(2) = 12$ and $3g(2) = 12$, we find $f(2) - g(2) = 8$. Evaluating this at $x = 2$ gives $8 = 8 + 6 + D$, resulting in $D = -6$, so the difference function is $h(x) = f(x) - g(x) = x^3 + 3x - 6$.

We can now verify the truth of each statement by examining $h(x)$. For the first option, evaluating at $x = -2$ gives $h(-2) = (-2)^3 + 3(-2) - 6 = -20$, which confirms $g(-2) - f(-2) = 20$. Regarding the second option, since $h'(x) = 3x^2 + 3$ is always positive, the function $h(x)$ is strictly increasing. Within the interval $(-1, 2)$, the function spans from $h(-1) = -10$ to $h(2) = 8$. Consequently, the statement $|h(x)| < 8$ is false because the function values extend to $-10$. For the third statement, setting $|f'(x) - g'(x)| < 6$ yields $|3x^2 + 3| < 6$, which simplifies to $x^2 < 1$, accurately describing the interval $-1 < x < 1$. Finally, for the last statement, we observe $h(1) = 1 + 3 - 6 = -2$ and $h\left(\frac{3}{2}\right) = \frac{27}{8} + \frac{9}{2} - 6 = 1.875$. Since the function changes sign between $1$ and $\frac{3}{2}$, the Intermediate Value Theorem guarantees that there exists some $x_0 \in \left(1, \frac{3}{2}\right)$ such that $f(x_0) - g(x_0) = 0$.

Recognizing that the given function simplifies to $f(x)=-\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)$ is the key to evaluating the expression efficiently. Differentiating this form yields the first derivative $f'(x)=-\frac{1}{2}\sec^2\left(\frac{x}{2}-\frac{\pi}{8}\right)$, and applying the chain rule again results in the second derivative $f''(x)=-\frac{1}{2}\sec^2\left(\frac{x}{2}-\frac{\pi}{8}\right)\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)$.

Substituting $x=\frac{7\pi}{12}$ into the inner term yields $\frac{7\pi}{24}-\frac{\pi}{8}=\frac{\pi}{6}$. The function value becomes $f\left(\frac{7\pi}{12}\right)=-\tan\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}$. Evaluating the second derivative at this point gives $f''\left(\frac{7\pi}{12}\right)=-\frac{1}{2}\sec^2\left(\frac{\pi}{6}\right)\tan\left(\frac{\pi}{6}\right)$, which simplifies to $-\frac{1}{2} \cdot \frac{4}{3} \cdot \frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$. Multiplying the two values yields $\left(-\frac{1}{\sqrt{3}}\right)\left(-\frac{2}{3\sqrt{3}}\right)=\frac{2}{9}$.

The derivative of $f(x)$ is $f'(x) = 2 + \frac{1}{1+x^2}$, which remains between $2.1$ and $3$ for all $x \in [0, 3]$. Similarly, the derivative of $g(x)$ is calculated as $g'(x) = \frac{1}{\sqrt{1+x^2}}$, which ranges from $\frac{1}{\sqrt{10}}$ to $1$ over the same interval. Because $f'(x)$ is consistently greater than $g'(x)$ for every $x$ in the domain, $f(x)$ grows at a faster rate than $g(x)$.

Both functions start at the same value, $f(0) = g(0) = 0$, and since $f(x)$ increases more rapidly, it maintains a higher value than $g(x)$ throughout the interval. Both functions are monotonically increasing, meaning they reach their maximum values at the upper bound, $x = 3$. Evaluating these at the endpoint yields $f(3) = 6 + \tan^{-1} 3$ and $g(3) = \ln(3 + \sqrt{10})$. Since $6 + \tan^{-1} 3$ is numerically greater than $\ln(3 + \sqrt{10})$, the maximum value of $f(x)$ necessarily exceeds the maximum value of $g(x)$.

Repeated composition of linear fractional transformations leads to a sequence of rational functions where the coefficients evolve according to a clear pattern. Computing the first few iterations of $f^1(x) = \frac{3x+2}{2x+3}$ reveals how the coefficients change relative to one another. The second iteration $f^2(x)$ yields $\frac{13x+12}{12x+13}$, and continuing the process for $f^3(x)$ results in $\frac{63x+62}{62x+63}$.

Identifying the underlying progression for the $n$-th iteration $f^n(x) = \frac{a_n x + b_n}{b_n x + a_n}$, the sum of the coefficients $a_n + b_n$ follows the sequence $5^n$. For $n=1$, the sum is $3+2=5^1$, for $n=2$, it is $13+12=5^2$, and for $n=3$, it is $63+62=5^3$. Consequently, for the fifth iteration, the sum $a+b$ corresponds directly to $5^5$, which is 3125.

Calculating the derivative of $y=x^x$ involves taking the natural logarithm of both sides to get $\ln y = x \ln x$, followed by implicit differentiation. This process leads to $\frac{y'}{y} = \ln x + 1$, resulting in $y' = x^x(1 + \ln x)$. To find the second derivative $y''$, apply the product rule to $y' = x^x(1 + \ln x)$, which yields $y'' = x^x(1 + \ln x)^2 + x^x \cdot \frac{1}{x}$.

Substituting $x=2$ into these derivatives gives $y'(2) = 4(1 + \ln 2)$ and $y''(2) = 4(1 + \ln 2)^2 + 2$. The expression $y''(2) - 2y'(2)$ therefore becomes $4(1 + \ln 2)^2 + 2 - 8(1 + \ln 2)$. Expanding the squared binomial results in $4(1 + 2 \ln 2 + (\ln 2)^2) + 2 - 8 - 8 \ln 2$, which simplifies to $4 + 8 \ln 2 + 4(\ln 2)^2 + 2 - 8 - 8 \ln 2$. Combining the constant terms and eliminating the linear logarithmic terms leaves $4(\ln 2)^2 - 2$.

To determine the function $f(x)$ from the given functional equation $f(x+y)=f(x)+f(y)-1$, we first evaluate the function at zero by substituting $x=0$ and $y=0$, which yields $f(0)=f(0)+f(0)-1$, confirming that $f(0)=1$. We then utilize the definition of the derivative $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ and substitute the functional equation $f(x+h)=f(x)+f(h)-1$ into the numerator to obtain $\lim_{h \to 0} \frac{f(x)+f(h)-1-f(x)}{h}$, which simplifies to $\lim_{h \to 0} \frac{f(h)-1}{h}$. Since $f(0)=1$, this limit is identical to $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$, which is the definition of $f'(0)$. Given that $f'(0)=2$, it follows that $f'(x)=2$ for all $x$. Integrating $f'(x)=2$ gives $f(x)=2x+C$, and using the initial condition $f(0)=1$, we find that $C=1$, resulting in $f(x)=2x+1$. Evaluating this at $x=-2$ gives $f(-2)=2(-2)+1=-3$, so the absolute value $|f(-2)|$ is $3$.

Because $f(x)$ is a cubic polynomial, its successive derivatives follow a predictable pattern where the first derivative is quadratic, the second is linear, and the third is a constant. By treating the unknown terms $f'(1)$, $f''(2)$, and $f'''(3)$ as constants $a$, $b$, and $c$ respectively, we define the function as $f(x)=x^{3-}ax^{2+}bx-c$. Differentiating this expression gives $f'(x)=3x^{2-}2ax+b$, $f''(x)=6x-2a$, and $f'''(x)=6$.

Since $c$ is defined as $f'''(3)$, it follows immediately that $c=6$. Applying the defined points to the derivatives, we have $f''(2)=6(2)-2a=12-2a$, which equals $b$, and $f'(1)=3(1)^{2-}2a+b=3-2a+b$, which equals $a$. Solving the system $b=12-2a$ and $a=3-2a+b$ yields $a=3$ and $b=6$. Substituting these values into our expression gives $f(x)=x^{3-}3x^{2+}6x-6$, from which we calculate the values $f(0)=-6$, $f(1)=-2$, $f(2)=2$, and $f(3)=12$. Evaluating $2f(0)-f(1)+f(3)$ leads to $2(-6)-(-2)+12$, which simplifies to $2$, matching the value of $f(2)$.

The expression $f(x) = \sum_{k=1}^{10} k x^k$ can be simplified by evaluating $(1-x)^2 f(x)$, which is derived from the geometric series property where $(1-x)f(x) = x + x^2 + \dots + x^{10} - 10x^{11}$. Rewriting this as $(1-x)^2 f(x) = x - x^{11} - 10x^{11}(1-x)$ and differentiating with respect to $x$ yields $(1-x)^2 f'(x) - 2(1-x)f(x) = 1 - 121x^{10} + 120x^{11}$. Evaluating this derivative at $x=2$ simplifies the expression to $f'(2) + 2f(2) = 1 - 121(2^{10}) + 240(2^{10}) = 1 + 119(2^{10})$. Because this equates to the form $119(2)^n + 1$, the value of $n$ is clearly $10$.

Evaluating the derivative of a logarithm with a variable base requires shifting to the natural logarithm using the change of base property, which allows the quotient rule to simplify the differentiation process. By rewriting $f(x) = \log_{\cos x} \csc x$ as $\frac{\ln(\csc x)}{\ln(\cos x)}$, the derivative is given by $\frac{-\cot x \ln(\cos x) + \tan x \ln(\csc x)}{(\ln(\cos x))^2}$ because the derivative of $\ln(\csc x)$ is $-\cot x$ and the derivative of $\ln(\cos x)$ is $-\tan x$. Substituting $x = \frac{\pi}{4}$ into this expression, where $\cot(\frac{\pi}{4}) = 1$, $\tan(\frac{\pi}{4}) = 1$, $\ln(\cos \frac{\pi}{4}) = -\frac{1}{2} \ln 2$, and $\ln(\csc \frac{\pi}{4}) = \frac{1}{2} \ln 2$, yields $\frac{-1(-\frac{1}{2} \ln 2) + 1(\frac{1}{2} \ln 2)}{(-\frac{1}{2} \ln 2)^2} = \frac{\ln 2}{\frac{1}{4}(\ln 2)^2} = \frac{4}{\ln 2}$. Multiplying this derivative by the factor $\log_e 2$, which is equivalent to $\ln 2$, results in a final value of 4.

The function $y(x)=(x^x)^x$ simplifies to $y=x^{x^2}$. Applying logarithmic differentiation by taking the natural logarithm on both sides yields $\ln y = x^2 \ln x$, and differentiating this equation with respect to $x$ leads to $\frac{1}{y} \frac{dy}{dx} = x + 2x \ln x$, which simplifies to the expression $\frac{dy}{dx} = x^{x^{2+}1}(1+2 \ln x)$. Since the second derivative $\frac{d^2 x}{d y^2}$ is equivalent to $\frac{d}{dx}\left(\frac{dx}{dy}\right) \frac{dx}{dy}$, it can be calculated by focusing on the reciprocal of the first derivative.

At $x=1$, the first derivative is $\frac{dy}{dx} = 1^{1^{2+}1}(1 + 2 \ln 1) = 1$, which implies $\frac{dx}{dy} = 1$. Determining the derivative of the reciprocal $\frac{dx}{dy}$ with respect to $x$ requires the quotient rule, and evaluating this derivative at $x=1$ yields a value of $-4$. Substituting these results into the formula $\frac{d^2 x}{d y^2} = \frac{d}{dx}\left(\frac{dx}{dy}\right) \frac{dx}{dy}$ produces $-4 \cdot 1 = -4$. Adding 20 to this value gives a final result of 16.

When a function $g(x)$ is the inverse of $f(x)$, satisfying $f(g(x)) = x$, their derivatives are linked by the chain rule applied to the composition, leading to the relationship $g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$. Given $f(x) = x^3 + x - 5$, we seek $g'(63)$, which requires finding the value of $x$ such that $f(x) = 63$. Setting $x^3 + x - 5 = 63$ results in $x^3 + x = 68$, which is satisfied when $x = 4$. Differentiating $f(x)$ yields $f'(x) = 3x^2 + 1$, so calculating $f'(4)$ gives $3(4)^2 + 1 = 49$. Substituting these values into the derivative relation provides $g'(63) = \frac{1}{49}$.