Differentiation – JEE Main MathematicsPractice Questions & PYQs

Substituting $x = y = \alpha$ into the curve equation $(x^{2+}y^{2-}3)+(x^{2-}y^{2-}1)^5=0$ allows us to solve for $\alpha$. This substitution yields $(\alpha^{2+}\alpha^{2-}3)+(\alpha^{2-}\alpha^{2-}1)^5=0$, which simplifies to $2\alpha^{2-}3-1=0$ and thus $2\alpha^2=4$. Given the condition $\alpha > 0$, we find $\alpha = \sqrt{2}$.

Differentiating the given curve equation with respect to $x$ involves applying the chain rule to obtain $2x+2yy'+5(x^{2-}y^{2-}1)^4(2x-2yy')=0$. Dividing by 2, we have $x+yy'+5(x^{2-}y^{2-}1)^4(x-yy')=0$. Evaluating this derivative at $x=y=\sqrt{2}$ gives $\sqrt{2}+\sqrt{2}y'+5(-1)^4(\sqrt{2}-\sqrt{2}y')=0$, which simplifies to $\sqrt{2}+\sqrt{2}y'+5\sqrt{2}-5\sqrt{2}y'=0$, resulting in $4\sqrt{2}y'=6\sqrt{2}$, or $y' = \frac{3}{2}$.

Differentiating the expression $x+yy'+5(x^{2-}y^{2-}1)^4(x-yy')=0$ once more with respect to $x$ leads to the second derivative. By substituting $x=y=\sqrt{2}$ and $y'=\frac{3}{2}$ into this result, we determine $y''=-\frac{23}{4\sqrt{2}}$. Substituting these values into the required expression $3y'-y^3y''$ gives $3(\frac{3}{2})-(\sqrt{2})^3(-\frac{23}{4\sqrt{2}})$. This simplifies to $\frac{9}{2} + 2\sqrt{2}(\frac{23}{4\sqrt{2}})$, which equals 4.5 + 11.5, resulting in a final value of 16.

The expression $f(x)g''(x) + f'(x)g'(x)$ is the derivative of the product function $h(x) = f(x)g'(x)$. Given that $f(x)$ is an even function with known zeros at $x = \frac{1}{4}$ and $x = \frac{1}{2}$, it must also possess zeros at $x = -\frac{1}{4}$ and $x = -\frac{1}{2}$ due to symmetry. Furthermore, since $g(x)$ is an even function, its derivative $g'(x)$ is an odd function, which necessitates that $g'(0) = 0$.

Combining these observations, the function $h(x) = f(x)g'(x)$ vanishes at the five distinct points $x \in \left\{ -\frac{1}{2}, -\frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2} \right\}$. According to Rolle's theorem, between every two consecutive zeros of a differentiable function, there exists at least one zero of its derivative. Therefore, the function $h'(x)$, which corresponds to the original expression $f(x)g''(x) + f'(x)g'(x)$, must have at least $5 - 1 = 4$ zeros in the interval $(-2, 2)$.

Recognizing that the composition can be simplified using trigonometric substitution makes the derivative calculation straightforward. By setting $2^x = \tan \theta$, the term inside the inverse cosine becomes $\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$, which is equivalent to $\cos 2\theta$. Since $f(x) = \sin(\cos^{-1}(\cos 2\theta))$, the function reduces to $\sin 2\theta$. We can express this in terms of $x$ using the identity $\sin 2\theta = \frac{2 \tan \theta}{1+\tan^2 \theta}$, which simplifies the original function to $f(x) = \frac{2 \cdot 2^x}{1+2^{2x}}$.

Applying the quotient rule to differentiate $f(x) = 2 \cdot \frac{2^x}{1+2^{2x}}$ yields $f'(x) = 2 \cdot \frac{(1+2^{2x})(2^x \ln 2) - 2^x(2^{2x} \ln 2 \cdot 2)}{(1+2^{2x})^2}$. Evaluating this derivative at $x=1$, where $2^x = 2$ and $2^{2x} = 4$, gives $f'(1) = 2 \cdot \frac{(5 \cdot 2 \ln 2 - 16 \ln 2)}{25} = -\frac{12}{25} \ln 2$. Comparing this result to $-\frac{b}{a} \ln 2$, we identify $a=25$ and $b=12$, which determines the final value as $|a^2 - b^2| = |25^2 - 12^2| = 481$.

The core principle here is to simplify the trigonometric terms under the radicals by recognizing them as perfect squares, specifically $1 \pm \sin x = (\cos(x/2) \pm \sin(x/2))^2$. Given the domain $x \in \left(\frac{\pi}{2}, \pi\right)$, we know that $\frac{x}{2} \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, a range where both sine and cosine are positive, but $\sin(x/2) > \cos(x/2)$. This inequality is critical for correctly removing the absolute values when taking the square roots.

Substituting the simplified squares into the original expression results in $\cot^{-1}\left(\frac{\cos(x/2)+\sin(x/2)+\sin(x/2)-\cos(x/2)}{\cos(x/2)+\sin(x/2)-(\sin(x/2)-\cos(x/2))}\right)$. After canceling out terms in the numerator and denominator, the expression reduces to $\cot^{-1}\left(\frac{2\sin(x/2)}{2\cos(x/2)}\right)$, which simplifies further to $\cot^{-1}(\tan(x/2))$. By applying the co-function identity $\tan(\theta) = \cot\left(\frac{\pi}{2}-\theta\right)$, we rewrite the entire function as $y(x) = \cot^{-1}\left(\cot\left(\frac{\pi}{2}-\frac{x}{2}\right)\right)$. This allows us to simplify the function to $y(x) = \frac{\pi}{2}-\frac{x}{2}$, and the derivative of this expression with respect to $x$ is $-\frac{1}{2}$.

Simplifying complex compositions of inverse trigonometric functions relies on working from the innermost expression outward. By identifying the innermost term as $\theta = \cot^{-1}\sqrt{\frac{1-x}{x}}$, we determine that $\tan\theta = \frac{\sqrt{x}}{\sqrt{1-x}}$, which corresponds to a right triangle with an opposite side of $\sqrt{x}$ and a hypotenuse of 1, confirming that $\theta = \sin^{-1}\sqrt{x}$. This simplification allows the function to collapse into $f(x) = \cos(2\tan^{-1}\sqrt{x})$, and by applying the double angle formula for cosine in terms of tangent, we find $f(x) = \frac{1-x}{1+x}$.

Given $f(x) = \frac{1-x}{1+x}$, calculating the derivative via the quotient rule yields $f'(x) = \frac{-(1+x)-(1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2}$. Substituting these results into the expression $(1-x)^2 f'(x) + 2(f(x))^2$ leads to $(1-x)^2 \left( \frac{-2}{(1+x)^2} \right) + 2 \left( \frac{1-x}{1+x} \right)^2$, which simplifies directly to $-2 \frac{(1-x)^2}{(1+x)^2} + 2 \frac{(1-x)^2}{(1+x)^2}$, resulting in zero.

Implicit differentiation involves treating $y$ as a function of $x$ and applying the chain rule consistently across both sides of the equation $\log_{e}(x+y)=4xy$. Setting $x=0$ reveals $y=1$ since $\log_{e}(1)=0$. Differentiating the equation with respect to $x$ yields $\frac{1+y'}{x+y} = 4y+4xy'$, which simplifies to $1+y' = (x+y)(4y+4xy')$. Substituting $x=0$ and $y=1$ into this expression shows $1+y' = (1)(4)$, confirming that the first derivative $y'$ is equal to $3$ at the origin.

Applying the product rule to differentiate $1+y' = (4y+4xy')e^{4xy}$ leads to the second derivative expression $y'' = e^{4xy}(4y+4xy')^2 + e^{4xy}(8y'+4xy'')$. Evaluating this at $x=0$ with the known values $y=1$ and $y'=3$ gives $1(4)^2 + 1(8 \times 3 + 0)$, which calculates to $16 + 24$. The final result for the second derivative $\frac{d^{2}y}{dx^{2}}$ is $40$.

Evaluating the equation $y^2 + \log_e(\cos^2 x) = y$ at $x=0$ reveals that since $\log_e(1) = 0$, the expression simplifies to $y^2 - y = 0$, meaning $y$ is either $0$ or $1$. Differentiating both sides with respect to $x$ yields $2yy' - 2\tan x = y'$, and setting $x=0$ shows that the first derivative $y'(0) = 0$ regardless of the value of $y$. Performing a second differentiation results in $2(y')^2 + 2yy'' - 2\sec^2 x = y''$, which simplifies at $x=0$ to $2yy'' - 2 = y''$ since $\sec^2(0) = 1$ and $y'(0) = 0$. Rearranging this leads to $y''(2y - 1) = 2$, where $y=0$ gives $y'' = -2$ and $y=1$ gives $y'' = 2$, confirming that the absolute value $|y''(0)| = 2$.

To determine $f(x)$, we let $\theta = \tan^{-1}x$, which implies $\cot^{-1}x = \frac{\pi}{2} - \theta$. The expression $f(x) = (\sin \theta + \cos \theta)^2 - 1$ simplifies using the identity $\sin^2 \theta + \cos^2 \theta = 1$, resulting in $1 + 2 \sin \theta \cos \theta - 1$, which is $\sin 2\theta$. Given $\tan \theta = x$, we can express this in terms of $x$ as $f(x) = \frac{2x}{1+x^2}$.

The given derivative condition is $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\sin^{-1}(\frac{2x}{1+x^2}))$. For $|x| > 1$, the derivative of $\sin^{-1}(\frac{2x}{1+x^2})$ is $-\frac{2}{1+x^2}$. Substituting this into the equation gives $\frac{dy}{dx} = \frac{1}{2} \left(-\frac{2}{1+x^2}\right) = -\frac{1}{1+x^2}$. Integrating this derivative results in $y = -\tan^{-1}x + C$, which is equivalent to $y = \cot^{-1}x + C_0$ for some constant $C_0$.

Using the condition $y(\sqrt{3}) = \frac{\pi}{6}$, we solve for the constant: $\cot^{-1}(\sqrt{3}) + C_0 = \frac{\pi}{6} + C_0 = \frac{\pi}{6}$, which implies $C_0 = 0$. With the function established as $y = \cot^{-1}x$, we calculate the value at $x = -\sqrt{3}$ as $y(-\sqrt{3}) = \cot^{-1}(-\sqrt{3})$. Since $\cot^{-1}(-a) = \pi - \cot^{-1}(a)$ for positive $a$, this is equal to $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

When two differentiable functions compose to yield the identity function, they function as inverses of each other, establishing a reciprocal relationship between their derivatives. This principle arises from the chain rule, where the product of the derivative of the outer function at the inner function's output and the derivative of the inner function equals the derivative of the identity map, which is 1.

Starting with $f(g(x)) = x$, differentiating both sides with respect to $x$ results in $f'(g(x)) \cdot g'(x) = 1$. Substituting $x = a$ transforms this into $f'(g(a)) \cdot g'(a) = 1$. Since $g(a) = b$ and $g'(a) = 5$, the relationship becomes $f'(b) \cdot 5 = 1$, which simplifies to $f'(b) = \frac{1}{5}$.

Simplifying complex trigonometric expressions usually relies on converting terms into sine and cosine or identifying algebraic patterns within the ratios. Looking at the expression inside the square root, we can rewrite $\tan\alpha + \cot\alpha$ as $\frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\sin\alpha}$, which simplifies to $\frac{1}{\sin\alpha\cos\alpha}$. Dividing this result by $1+\tan^2\alpha$, which is equivalent to $\sec^2\alpha$, yields $\frac{\cos\alpha}{\sin\alpha}$, or $\cot\alpha$. Incorporating this into the full expression produces $2\cot\alpha + \csc^2\alpha$, and since $1+\cot^2\alpha = \csc^2\alpha$, this expression becomes $\cot^2\alpha + 2\cot\alpha + 1$, forming the perfect square $(\cot\alpha + 1)^2$.

Because $\alpha$ lies in the interval $\left(\frac{3\pi}{4}, \pi\right)$, the value of $\cot\alpha$ is less than $-1$, which makes the quantity $\cot\alpha + 1$ negative. Consequently, the square root of $(\cot\alpha + 1)^2$ is equal to $-(1+\cot\alpha)$, simplifying the function to $-1-\cot\alpha$. Differentiating this expression with respect to $\alpha$ yields $-\frac{d}{d\alpha}(\cot\alpha)$, which is $\csc^2\alpha$. Evaluating this result at $\alpha=\frac{5\pi}{6}$ requires finding the square of the cosecant function, where $\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$ leads to $\csc\left(\frac{5\pi}{6}\right) = 2$, ultimately resulting in 4.

The expression within the inverse cosine function can be simplified by recognizing that the coefficients $\frac{3}{5}$ and $\frac{4}{5}$ represent the cosine and sine values of a specific angle $\alpha$. By rewriting the inner term as $\cos \alpha \cos kx - \sin \alpha \sin kx$, we effectively collapse it into the trigonometric addition identity $\cos(\alpha + kx)$. Applying the identity $\cos^{-1}(\cos \theta) = \theta$ for the relevant range near zero, the function simplifies to $y = \sum_{k=1}^{6} k(\alpha + kx)$, which expands to $\sum_{k=1}^{6} (k\alpha + k^2x)$.

Differentiating this expression with respect to $x$ causes the constant terms containing $\alpha$ to vanish, leaving only the derivative of the $k^2x$ terms, which is $\sum_{k=1}^{6} k^2$. Calculating this sum of squares using the formula $\frac{n(n+1)(2n+1)}{6}$ with $n=6$, we compute $\frac{6 \times 7 \times 13}{6}$, which yields 91.

The equation $y\sqrt{1-x^2} = k - x\sqrt{1-y^2}$ reflects the trigonometric identity for the sine of a sum, $\sin(\alpha + \theta) = \sin \alpha \cos \theta + \cos \alpha \sin \theta$. By setting $x = \sin \theta$ and $y = \sin \alpha$, the given relationship simplifies to $\sin(\alpha + \theta) = k$, which implies the constant sum $\arcsin(y) + \arcsin(x) = \arcsin(k)$.

Differentiating both sides of the relation with respect to $x$ yields $\frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} = 0$, which allows us to define the derivative as $\frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$. Substituting the given values $x = \frac{1}{2}$ and $y = -\frac{1}{4}$ results in $\sqrt{1-x^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ and $\sqrt{1-y^2} = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$. Combining these values into the derivative expression, we obtain $-\frac{\sqrt{15}/4}{\sqrt{3}/2} = -\frac{\sqrt{5}}{2}$.

Trigonometric substitution is the most effective approach for simplifying complex inverse tangent expressions. For the first function, substituting $x = \tan \theta$ transforms the argument into
$\tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) = \frac{\theta}{2}$
which equals $\frac{1}{2} \tan^{-1} x$, yielding a derivative with respect to $x$ of $\frac{1}{2(1+x^2)}$.

For the second function, using the substitution $x = \sin \theta$ leads to
$\tan^{-1}\left(\frac{2 \sin \theta \cos \theta}{1 - 2 \sin^2 \theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta$
or $2 \sin^{-1} x$, which has a derivative with respect to $x$ of $\frac{2}{\sqrt{1-x^2}}$. Dividing the first derivative by the second results in the expression $\frac{\sqrt{1-x^2}}{4(1+x^2)}$, and evaluating this at $x = \frac{1}{2}$ produces $\frac{\sqrt{3}}{10}$.

The second derivative of the parametric functions $x = 2\sin\theta - \sin 2\theta$ and $y = 2\cos\theta - \cos 2\theta$ is obtained by applying the chain rule to the first derivative, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. Differentiating the given functions with respect to $\theta$ yields $\frac{dx}{d\theta} = 2\cos\theta - 2\cos 2\theta$ and $\frac{dy}{d\theta} = -2\sin\theta + 2\sin 2\theta$. By applying trigonometric sum-to-product identities, we can simplify $\frac{dy}{d\theta}$ to $4\sin(\theta/2)\cos(3\theta/2)$ and $\frac{dx}{d\theta}$ to $4\sin(\theta/2)\sin(3\theta/2)$, which simplifies the ratio $\frac{dy}{dx}$ to $\cot(3\theta/2)$.

To find the second derivative, $\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\cot\left(\frac{3\theta}{2}\right)\right) \cdot \frac{d\theta}{dx}$, we calculate the derivative of $\cot(3\theta/2)$ as $-\frac{3}{2}\csc^2(3\theta/2)$. Multiplying this by the reciprocal of $\frac{dx}{d\theta}$, given by $\frac{1}{4\sin(\theta/2)\sin(3\theta/2)}$, provides the full expression for $\frac{d^2y}{dx^2}$. Evaluating this expression at $\theta = \pi$ involves substituting $\csc^2(3\pi/2) = (-1)^2 = 1$, $\sin(\pi/2) = 1$, and $\sin(3\pi/2) = -1$, which leads to the final calculation of $\left(-\frac{3}{2}\right) \cdot (1) \cdot \frac{1}{4 \cdot 1 \cdot (-1)}$, resulting in a value of $\frac{3}{8}$.

The relationship between the variables $x$ and $y$ governed by the equation $x^k+y^k=a^k$ is best analyzed through implicit differentiation with respect to $x$. Performing this differentiation on both sides of the equation leads to the result:
$kx^{k-1}+ky^{k-1}\frac{dy}{dx}=0$
Isolating the derivative yields $\frac{dy}{dx}=-\frac{x^{k-1}}{y^{k-1}}$, which simplifies to $-\left(\frac{y}{x}\right)^{-(k-1)}$. By comparing this expression to the provided differential equation $\frac{dy}{dx}=-\left(\frac{y}{x}\right)^{1/3}$, we can set the exponents equal such that $-(k-1)=\frac{1}{3}$. Solving this equation leads directly to $k-1=-\frac{1}{3}$, resulting in $k=\frac{2}{3}$.

When dealing with functions defined parametrically, finding the second derivative requires calculating the derivative of the first slope with respect to the parameter $t$ and adjusting it using the chain rule. We first determine the individual derivatives $\frac{dx}{dt} = 3 \sec^2 t$ and $\frac{dy}{dt} = 3 \sec t \tan t$, which allow us to find the first derivative $\frac{dy}{dx}$ by taking the ratio $\frac{dy/dt}{dx/dt}$, resulting in $\frac{3 \sec t \tan t}{3 \sec^2 t} = \sin t$.

To obtain the second derivative $\frac{d^2y}{dx^2}$, we differentiate $\sin t$ with respect to $t$ to get $\cos t$, and then multiply this result by $\frac{dt}{dx}$, which is the reciprocal of $\frac{dx}{dt}$, yielding $\cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos^3 t}{3}$. Substituting $t = \frac{\pi}{4}$ into this expression, where $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, leads to the final calculation of $\frac{(1/\sqrt{2})^3}{3}$, which simplifies to $\frac{1}{6\sqrt{2}}$.

Applying the natural logarithm to both sides of the equation $(2x)^{2y} = 4e^{2x-2y}$ is the most effective approach here, as it converts variable exponents into a linear relationship that can be differentiated more easily. Taking the logarithm on both sides results in $2y \log_e 2x = \log_e 4 + 2x - 2y$, which simplifies the expression by moving the variables out of the exponents. Rearranging this equation to solve for $y$ yields $2y(\log_e 2x + 1) = 2x + 2\log_e 2$, and dividing through gives $y = \frac{x+\log_e 2}{1+\log_e 2x}$.

Differentiating this function with respect to $x$ using the quotient rule, we obtain $\frac{dy}{dx} = \frac{(1+\log_e 2x)(1) - (x+\log_e 2)(\frac{1}{x})}{(1+\log_e 2x)^2}$. Multiplying this derivative by the factor $(1+\log_e 2x)^2$ cancels the denominator and isolates the numerator, which simplifies to $1 + \log_e 2x - 1 - \frac{\log_e 2}{x}$, resulting in the final expression $\frac{x\log_e 2x - \log_e 2}{x}$.

The composition $(f \circ g)(x)$ simplifies directly to $-x$ because the natural logarithm and the exponential function are inverses of one another, effectively cancelling out once the sine and inverse sine operations are applied. Evaluating this result at $x = \alpha$ gives $b = -\alpha$, and differentiating with respect to $x$ reveals the constant value $a = -1$. Substituting these findings into the expression $a\alpha^{2}-b\alpha-a$ results in $(-1)\alpha^2 - (-\alpha)\alpha - (-1)$, which simplifies perfectly to $-\alpha^2 + \alpha^2 + 1 = 1$.