Differentiation – JEE Main MathematicsPractice Questions & PYQs

Simplifying the first term by dividing the numerator and denominator by $4\cos x$ results in $\tan^{-1}\left(\frac{3/4 - \tan x}{1 + (3/4)\tan x}\right)$, which simplifies to $\tan^{-1}(3/4) - x$ using the difference of two angles formula. For the second term, setting $x = \sin\theta$ reduces the expression $2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ to $2\tan^{-1}(\tan(\theta/2))$, resulting in $\sin^{-1}x$. With the full function $y = \tan^{-1}(3/4) - x + \sin^{-1}x$, the derivative $\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-x^2}}$ at $x = \frac{\sqrt{3}}{2}$ becomes $-1 + \frac{1}{\sqrt{1 - 3/4}}$, which equals $1$.

Given the polynomial $f(x)=x^{3+}x^2 f'(1)+2 x f''(2)+f'''(3)$, we can identify the constant coefficients by letting $a=f'(1)$, $b=f''(2)$, and $c=f'''(3)$. Since $f(x)=x^{3+}ax^{2+}2bx+c$, we have $f'(x)=3x^{2+}2ax+2b$, $f''(x)=6x+2a$, and $f'''(x)=6$, which implies $c=6$. Applying the defined constants to these derivatives, we obtain the system of equations $a=f'(1)=3(1)^{2+}2a(1)+2b$ and $b=f''(2)=6(2)+2a$.

Simplifying these equations leads to $a+2b=-3$ and $2a-b=-12$. Substituting $b=2a+12$ into the first equation yields $a+2(2a+12)=-3$, which simplifies to $5a=-27$ or $a=-27/5$. Substituting this value back gives $b=2(-27/5)+12 = -54/5 + 60/5 = 6/5$. Finally, we evaluate $f'(x)=3x^{2+}2ax+2b$ at $x=5$: $f'(5)=3(5^2)+2(-27/5)(5)+2(6/5) = 75-54+12/5 = 21+12/5 = 117/5$.

Expanding the determinant along the first row allows for the simplification of the trigonometric expression into a manageable form. Applying this expansion yields $\sin x(28 - 27) - \cos x(27 - 27) + (\sin x + \cos x + 1)(27 - 28)$, which reduces directly to $\sin x(1) - \cos x(0) - (\sin x + \cos x + 1)$. Simplifying this algebraic expression results in $y(x) = -\cos x - 1$.

Calculating the derivatives of this function provides $y'(x) = \sin x$ and $y''(x) = \cos x$. Substituting the second derivative and the original function into the expression $\frac{d^2 y}{d x^2} + y$ gives $\cos x + (-\cos x - 1)$, which results in $-1$.

The differential equation $f''(x) = f(x)$ indicates that the function $f(x)$ takes the general form $A e^x + B e^{-x}$ for constants $A$ and $B$. Given $f(0) = 0$, the relationship $A + B = 0$ must hold, which implies $B = -A$ and leads to the expression $f(x) = A(e^x - e^{-x})$. Differentiating this function yields $f'(x) = A(e^x + e^{-x})$, and the condition $f'(0) = 3$ establishes that $A(e^0 + e^{-0}) = 3$, resulting in $2A = 3$, or $A = 3/2$.

Substituting $x = \ln 3$ into the function $f(x) = \frac{3}{2}(e^x - e^{-x})$ gives $f(\ln 3) = \frac{3}{2}(e^{\ln 3} - e^{-\ln 3})$. Using the properties $e^{\ln 3} = 3$ and $e^{-\ln 3} = 1/3$, the value becomes $\frac{3}{2}(3 - 1/3)$, which simplifies to $\frac{3}{2} \times \frac{8}{3} = 4$. Consequently, the final value of $9 f(\ln 3)$ is $9 \times 4 = 36$.

Expanding the product terms of the functional identity reveals the hidden trigonometric addition and subtraction formulas. Distributing the terms on both sides of the equation leads to:
$f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\sin x \cos y + \cos x \sin y)$
Applying the sine sum and difference identities reduces this to $f(2x+2y) \sin(x-y) = f(2x-2y) \sin(x+y)$. Rearranging this equality yields the ratio $\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$, which indicates that the expression $\frac{f(t)}{\sin(t/2)}$ must equal a constant $K$, identifying the function as $f(x) = K \sin(\frac{x}{2})$.

Applying the slope condition $f^{\prime}(0) = \frac{1}{2}$ to the derivative $f^{\prime}(x) = \frac{K}{2} \cos(\frac{x}{2})$ results in $\frac{K}{2} \cos(0) = \frac{1}{2}$, which confirms $K=1$. Given $f(x) = \sin(\frac{x}{2})$, the second derivative is $f^{\prime\prime}(x) = -\frac{1}{4} \sin(\frac{x}{2})$. Evaluating this at $\frac{5\pi}{3}$ gives $f^{\prime\prime}(\frac{5\pi}{3}) = -\frac{1}{4} \sin(\frac{5\pi}{6}) = -\frac{1}{4} \times \frac{1}{2} = -\frac{1}{8}$. Therefore, the value of $24 f^{\prime\prime}(\frac{5\pi}{3})$ is $24 \times (-\frac{1}{8}) = -3$.

The core principle for solving this problem is identifying the expression as a first-order linear ordinary differential equation, which takes the general form $\frac{dy}{dx} + P(x)y = Q(x)$. By dividing the given equation $x^2 f'(x) = 2 x f(x) + 3$ by $x^2$, we rewrite it as $f'(x) - \frac{2}{x} f(x) = \frac{3}{x^2}$. The integrating factor required to solve this is $e^{\int -\frac{2}{x} dx}$, which simplifies to $e^{-2 \ln x} = \frac{1}{x^2}$. Multiplying the differential equation by this factor yields the derivative of the product, $\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = \frac{3}{x^4}$. Integrating both sides with respect to $x$ gives $\frac{f(x)}{x^2} = \int 3x^{-4} dx = -x^{-3} + C$, which simplifies to $f(x) = -\frac{1}{x} + Cx^2$. Applying the initial condition $f(1) = 4$ results in $4 = -1 + C$, confirming that $C = 5$ and defining the function as $f(x) = -\frac{1}{x} + 5x^2$. Evaluating $2 f(2)$ involves calculating $2 \left( -\frac{1}{2} + 5(2)^2 \right)$, which simplifies to $2 \left( -\frac{1}{2} + 20 \right) = -1 + 40 = 39$.

The functional equation $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ implies a specific derivative relationship, which is found by differentiating with respect to $x$ while holding $y$ constant to obtain $f'\left(\frac{x}{y}\right) \cdot \frac{1}{y} = \frac{f'(x)}{f(y)}$. Setting $y = x$ transforms this equality into $f'(1) \cdot \frac{1}{x} = \frac{f'(x)}{f(x)}$, and by substituting the given $f'(1) = 2024$, we arrive at $\frac{2024}{x} = \frac{f'(x)}{f(x)}$. Rearranging this final expression by multiplying across results in the differential equation $x f'(x) - 2024 f(x) = 0$.

The key to solving calculus problems involving complex expressions is to prioritize simplification before attempting differentiation. For the first term, observing the algebraic structure allows us to factor the numerator and denominator. The denominator can be written as $\sqrt{x}(x + \sqrt{x} + 1)$, while the numerator factors as $(\sqrt{x}+1) \cdot \sqrt{x} \cdot ((\sqrt{x})^3 - 1)$. Using the difference of cubes identity, $((\sqrt{x})^3 - 1)$ becomes $(\sqrt{x} - 1)(x + \sqrt{x} + 1)$. After canceling the common factor $(x + \sqrt{x} + 1)$ and $\sqrt{x}$, the first part of the expression simplifies cleanly to $(\sqrt{x}+1)(\sqrt{x}-1)$, which is simply $x-1$.

The second term, $\frac{1}{15}(3\cos^2 x - 5)\cos^3 x$, expands to $\frac{1}{5}\cos^5 x - \frac{1}{3}\cos^3 x$. With the expression now reduced to $y = x - 1 + \frac{1}{5}\cos^5 x - \frac{1}{3}\cos^3 x$, finding the derivative $y'$ becomes a straightforward application of the chain rule. The derivative is $y' = 1 + \frac{1}{5}(5\cos^4 x)(-\sin x) - \frac{1}{3}(3\cos^2 x)(-\sin x)$, which simplifies to $y' = 1 - \cos^4 x \sin x + \cos^2 x \sin x$.

Evaluating this derivative at $x = \frac{\pi}{6}$ requires substituting $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. This yields $y'(\frac{\pi}{6}) = 1 - (\frac{3}{4})^2(\frac{1}{2}) + (\frac{3}{4})(\frac{1}{2})$, which further simplifies to $1 - \frac{9}{16} \cdot \frac{1}{2} + \frac{3}{8}$. Calculating the fraction gives $1 - \frac{9}{32} + \frac{12}{32} = 1 + \frac{3}{32} = \frac{35}{32}$. Multiplying this result by $96$ gives $96 \cdot \frac{35}{32}$, resulting in $3 \cdot 35 = 105$.

Determining the coefficients of the cubic polynomial $f(x)=a x^{3+}b x^{2+}c x+41$ involves utilizing its derivatives, where $f'(x)=3a x^{2+}2b x+c$ and $f''(x)=6a x+2b$. Evaluating these at $x=1$ yields the equations $3a+2b+c=2$ and $6a+2b=4$. Simplifying the second equation provides $3a+b=2$, and subtracting this from the first equation reveals $b+c=0$. Using the condition $f(1)=a+b+c+41=40$, we obtain $a+b+c=-1$, which simplifies to $a=-1$ after substituting $b+c=0$. Replacing $a=-1$ in the equation $3a+b=2$ gives $b=5$, which further implies $c=-5$ from the relation $b+c=0$. The value of $a^{2+}b^{2+}c^2$ is therefore $(-1)^{2+}5^{2+}(-5)^2=1+25+25=51$.

Finding the derivative of the function $g(x) = h(e^x) e^{h(x)}$ relies on the product rule, which states that the derivative of a product of two functions is the derivative of the first multiplied by the second, plus the first multiplied by the derivative of the second. Applying this along with the chain rule leads to the expression g'(x) = \[h'(e^x) \cdot e^x]$ \cdot e^{h(x)} + h(e^x) \cdot [e^{h(x)} \cdot h'(x)]$.

Evaluating this derivative at $x=0$ requires substituting the given values $h(0) = 0$, $h(1) = 1$, $h'(0) = 2$, and $h'(1) = 2$, while utilizing the fact that $e^0 = 1$. Substituting these values into the derived expression results in $g'(0) = [h'(1) \cdot 1] \cdot e^{h(0)} + h(1) \cdot [e^{h(0)} \cdot h'(0)]$. Inserting the specific numbers produces $g'(0) = [2 \cdot 1] \cdot 1 + 1 \cdot [1 \cdot 2]$, which simplifies to $2 + 2$, giving a final value of 4.

The function $f(x)=x^{3+}x^2 f'(1)+x f''(2)+f'''(3)$ for $x \in R$ involves coefficients that are derivatives of the function itself, which we evaluate through repeated differentiation: $f'(x)=3x^{2+}2x f'(1)+f''(2)$, $f''(x)=6x+2 f'(1)$, and $f'''(x)=6$. Given that $f'''(x)=6$, we identify $f'''(3)=6$ and substitute this into the expression for the second derivative at $x=2$ to obtain $f''(2)=6(2)+2 f'(1)$, or $f''(2)=12+2 f'(1)$. Substituting these results into the definition $f'(1)=3(1)^{2+}2(1) f'(1)+f''(2)$ creates the equation $f'(1)=3+2 f'(1)+12+2 f'(1)$, which simplifies to $-3 f'(1)=15$, meaning $f'(1)=-5$. With $f'(1)=-5$, we determine $f''(2)=12+2(-5)=2$, allowing us to finalize the derivative formula as $f'(x)=3x^{2-}10x+2$, so that evaluating at $x=10$ results in $f'(10)=3(100)-10(10)+2=202$.

By differentiating the definition $f(x) = \frac{1}{2}[g(x) + g(2-x)]$ with respect to $x$, we arrive at $f'(x) = \frac{g'(x) - g'(2-x)}{2}$. Substituting $x = \frac{1}{2}$ yields $f'(\frac{1}{2}) = \frac{g'(\frac{1}{2}) - g'(\frac{3}{2})}{2}$, which equals $0$ since $g'(\frac{1}{2}) = g'(\frac{3}{2})$. Similarly, substituting $x = \frac{3}{2}$ produces $f'(\frac{3}{2}) = \frac{g'(\frac{3}{2}) - g'(\frac{1}{2})}{2}$, which also results in $0$. Given that both $\frac{1}{2}$ and $\frac{3}{2}$ reside within the interval $(0, 2)$, the derivative $f'(x)$ vanishes at these two distinct points, confirming that it is zero at least twice in that range.

The function $\frac{2\cos\theta+\cos 2\theta}{\cos 3\theta+4\cos 2\theta+5\cos\theta+2}$ simplifies significantly by applying standard trigonometric identities. Expanding $\cos 2\theta$ as $2\cos^2\theta - 1$ and $\cos 3\theta$ as $4\cos^3\theta - 3\cos\theta$ allows the entire expression to be rewritten as a ratio containing the common factor $(2\cos^2\theta + 2\cos\theta - 1)$ in both the numerator and the denominator. After cancelling this common term, the function reduces to $y = \frac{1}{2(1+\cos\theta)}$.

Substituting $\theta = \frac{\pi}{2}$ into this simplified expression gives $y = \frac{1}{2(1+0)} = \frac{1}{2}$. Differentiating $y$ with respect to $\theta$ yields $y' = \frac{\sin\theta}{2(1+\cos\theta)^2}$, which evaluates to $\frac{1}{2(1)^2} = \frac{1}{2}$ at the given angle. Performing a second differentiation produces $y'' = \frac{\cos\theta(1+\cos\theta)^2 + 2\sin^2\theta(1+\cos\theta)}{2(1+\cos\theta)^4}$, which simplifies to $1$ at $\theta = \frac{\pi}{2}$. Adding these individual results together gives $1+\frac{1}{2}+\frac{1}{2} = 2$.

The given limit expression $\lim_{a\to \infty } \frac{a(a+1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2 \ln a$ represents the derivative $f'(1)$ of a function $f(x)$ that is defined by the transformation $x = \frac{1}{a}$. By substituting $x = \frac{1}{a}$ into the expression and accounting for the scaling factor, we identify the function as $f(x) = \frac{1+x}{2} \tan^{-1}(x) + 1 + 2x^2 \ln x$. Differentiating this function with respect to $x$ using the product and chain rules yields $f'(x) = \frac{1}{2} \left( \tan^{-1}(x) + \frac{1+x}{1+x^2} \right) + 4x \ln x + 2x$. Evaluating this derivative at $x=1$, where $\tan^{-1}(1) = \frac{\pi}{4}$ and $\ln(1) = 0$, simplifies to $f'(1) = \frac{1}{2} \left( \frac{\pi}{4} + 1 \right) + 0 + 2$. Combining these terms gives the final value of $\frac{5}{2} + \frac{\pi}{8}$.

Starting with $\log_e y = 3 \sin^{-1} x$, differentiating both sides gives $\frac{y'}{y} = \frac{3}{\sqrt{1-x^2}}$, which can be rearranged as $\sqrt{1-x^2} y' = 3y$. Differentiating this expression again with respect to $x$ using the product rule leads to $\sqrt{1-x^2} y'' - \frac{x y'}{\sqrt{1-x^2}} = 3y'$, and multiplying throughout by $\sqrt{1-x^2}$ yields the identity $(1-x^2) y'' - x y' = 3 y' \sqrt{1-x^2}$. Since the first derivative rearrangement establishes $\sqrt{1-x^2} y' = 3y$, the expression further reduces to $(1-x^2) y'' - x y' = 3(3y) = 9y$. Substituting $x = \frac{1}{2}$ into $y = e^{3 \sin^{-1} x}$ gives $y = e^{3(\pi/6)} = e^{\pi/2}$, so the final value is $9 e^{\pi/2}$.

Simplifying the expression $y = \log_e \left( \frac{1-x^2}{1+x^2} \right)$ using logarithmic properties as $y = \log_e(1-x^2) - \log_e(1+x^2)$ makes the differentiation process straightforward. Applying the derivative rules for logarithms, the first derivative is $y' = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2}$, which combines to $y' = \frac{-4x}{1-x^4}$.

To find the second derivative $y''$, differentiating the previous result using the quotient rule yields $y'' = \frac{-4(1+3x^4)}{(1-x^4)^2}$. Combining these expressions requires finding $y' - y'' = \frac{-4x}{1-x^4} + \frac{4(1+3x^4)}{(1-x^4)^2}$.

At the specific point $x = \frac{1}{2}$, substituting the value into the combined expression $y' - y''$ results in $\frac{736}{225}$. Scaling this value by 225 effectively isolates the numerator, resulting in 736.

The relationship $(g \circ f)(x) = x$ signifies that $g$ is the inverse of $f$, allowing us to use the chain rule to differentiate both sides as $g'(f(x)) \cdot f'(x) = 1$. To determine $g'(2)$, we first find the value of $x$ where $f(x) = 2$; observing that $f(0) = 0^5 + 2e^0 = 2$, we proceed to evaluate the derivative at $x = 0$. The derivative $f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$ at $x = 0$ simplifies to $f'(0) = 5(0)^4 + \frac{1}{2}e^0 = \frac{1}{2}$. Substituting this value into the inverse derivative identity gives $g'(2) = \frac{1}{f'(0)} = 2$. Consequently, calculating $8g'(2)$ leads to $8 \times 2 = 16$.

The definition of the derivative at a specific point provides a direct approach to evaluating $f'(0)$ when the function evaluates to 0 at the origin. By applying the definition $f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h}$, the expression simplifies significantly because $f(0) = \frac{(2^{0+}2^{-0}) \tan 0 \sqrt{\tan^{-1}(0^{2-}0+1)}}{(7(0)^{2+}3(0)+1)^3}$, which results in 0.

Substituting the expression for $f(h)$ into the limit yields $\lim_{h \to 0} \frac{(2^h+2^{-h}) \tan h \sqrt{\tan^{-1}(h^{2-}h+1)}}{h(7h^{2+}3h+1)^3}$. This can be evaluated by isolating the standard limit component $\frac{\tan h}{h}$, which approaches 1 as $h$ approaches 0. Simultaneously, the term $(2^h+2^{-h})$ approaches 2, the denominator $(7h^{2+}3h+1)^3$ approaches 1, and the radical component $\sqrt{\tan^{-1}(h^{2-}h+1)}$ approaches $\sqrt{\tan^{-1}(1)}$, which is equivalent to $\sqrt{\frac{\pi}{4}}$ or $\frac{\sqrt{\pi}}{2}$. Multiplying these limits as $2 \cdot 1 \cdot \frac{\sqrt{\pi}}{2}$ confirms that the final value is $\sqrt{\pi}$.

Recognizing that the expression $3 f' f'' + f f'''$ is equivalent to the second derivative of the product $f(x) f'(x)$ simplifies the problem significantly, as we can analyze the zeros of this product instead. The first derivative of $f(x) f'(x)$ is $(f'(x))^2 + f(x) f''(x)$, and its second derivative is $2 f'(x) f''(x) + f'(x) f''(x) + f(x) f'''(x)$, which directly simplifies to $3 f' f'' + f f'''$.

With $f(0)=0$ and the given values $f(1)=1, f(2)=-1, f(3)=2, f(4)=-2$, the intermediate value theorem implies that $f(x)$ must cross the x-axis at least once in each interval $(1, 2)$, $(2, 3)$, and $(3, 4)$, providing three roots in addition to the root at $x=0$. This gives $f(x)$ at least four distinct zeros. According to Rolle's theorem, there must be at least one root of $f'(x)$ between each adjacent pair of these four zeros, which ensures that $f'(x)$ has at least three distinct roots.