DERIVATIVES IMPORTANT QUESTIONS WITH SOLUTION FOR BOARD EXAM MAHARASHTRA STATE BOARD
MATHEMATICS
Problem 1 Solution (Image: 1.png)
Find $\frac{dy}{dx}$ for $y = \log(\sec x + \tan x)$.
Solution Steps:
Let $y = \log(\sec x + \tan x)$.
We use the chain rule. Let $u = \sec x + \tan x$. Then $y = \log u$.
The derivative of $\log u$ with respect to $u$ is $\frac{dy}{du} = \frac{1}{u}$.
The derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = \frac{d}{dx}(\sec x + \tan x) = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) $$
$$ \frac{du}{dx} = \sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x) $$
Now, apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:
$$ \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \sec x (\sec x + \tan x) $$
$$ \frac{dy}{dx} = \sec x $$
Final Answer:
$$ \frac{dy}{dx} = \sec x $$Problem 2 Solution (Image: 2.png, first)
Find $\frac{dy}{dx}$ for $y = \log(\tan(\frac{x}{2}))$.
Solution Steps:
Let $y = \log(\tan(\frac{x}{2}))$.
We use the chain rule. Let $u = \tan(\frac{x}{2})$ and $v = \frac{x}{2}$. Then $y = \log u$, $u = \tan v$.
$$ \frac{dy}{du} = \frac{1}{u} = \frac{1}{\tan(x/2)} $$
$$ \frac{du}{dv} = \sec^2 v = \sec^2(x/2) $$
$$ \frac{dv}{dx} = \frac{1}{2} $$
Using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$:
$$ \frac{dy}{dx} = \frac{1}{\tan(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} $$
$$ \frac{dy}{dx} = \frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{\cos^2(x/2)} \cdot \frac{1}{2} $$
$$ \frac{dy}{dx} = \frac{1}{2 \sin(x/2) \cos(x/2)} $$
Using the identity $\sin x = 2 \sin(x/2) \cos(x/2)$:
$$ \frac{dy}{dx} = \frac{1}{\sin x} = \csc x $$
Final Answer:
$$ \frac{dy}{dx} = \csc x $$Problem 3 Solution (Image: 3.png)
Find $\frac{dy}{dx}$ for $y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.
Solution Steps:
First, simplify the expression for $y$: $$ y = \log \left( \left( \frac{1 - \cos x}{1 + \cos x} \right)^{1/2} \right) = \frac{1}{2} \log \left( \frac{1 - \cos x}{1 + \cos x} \right) $$ Using half-angle identities $1 - \cos x = 2 \sin^2(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$: $$ y = \frac{1}{2} \log \left( \frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} \right) = \frac{1}{2} \log \left( \tan^2(x/2) \right) $$ Assuming $\tan(x/2) > 0$: $$ y = \frac{1}{2} \cdot 2 \log \left( \tan(x/2) \right) = \log \left( \tan(x/2) \right) $$ This is the same expression as in Problem 2. Therefore, the derivative is: $$ \frac{dy}{dx} = \csc x $$ Alternatively, differentiating $y = \frac{1}{2} (\log(1 - \cos x) - \log(1 + \cos x))$: $$ \frac{dy}{dx} = \frac{1}{2} \left( \frac{\sin x}{1 - \cos x} - \frac{-\sin x}{1 + \cos x} \right) = \frac{\sin x}{2} \left( \frac{1}{1 - \cos x} + \frac{1}{1 + \cos x} \right) $$ $$ \frac{dy}{dx} = \frac{\sin x}{2} \left( \frac{1 + \cos x + 1 - \cos x}{(1 - \cos x)(1 + \cos x)} \right) = \frac{\sin x}{2} \left( \frac{2}{1 - \cos^2 x} \right) $$ $$ \frac{dy}{dx} = \frac{\sin x}{2} \left( \frac{2}{\sin^2 x} \right) = \frac{1}{\sin x} = \csc x $$
Final Answer:
$$ \frac{dy}{dx} = \csc x $$Problem 4 Solution (Image: 4.png)
Find $\frac{dy}{dx}$ for $y = \log \sqrt{\frac{1 + \sin x}{1 - \sin x}}$.
Solution Steps:
First, simplify the expression for $y$: $$ y = \log \left( \left( \frac{1 + \sin x}{1 - \sin x} \right)^{1/2} \right) = \frac{1}{2} \log \left( \frac{1 + \sin x}{1 - \sin x} \right) $$ Rationalize the argument of the logarithm: $$ \frac{1 + \sin x}{1 - \sin x} = \frac{1 + \sin x}{1 - \sin x} \cdot \frac{1 + \sin x}{1 + \sin x} = \frac{(1 + \sin x)^2}{1 - \sin^2 x} = \frac{(1 + \sin x)^2}{\cos^2 x} = \left( \frac{1 + \sin x}{\cos x} \right)^2 = (\sec x + \tan x)^2 $$ So, $$ y = \frac{1}{2} \log ((\sec x + \tan x)^2) $$ Assuming $\sec x + \tan x > 0$: $$ y = \frac{1}{2} \cdot 2 \log (\sec x + \tan x) = \log (\sec x + \tan x) $$ This is the same expression as in Problem 1. Therefore, the derivative is: $$ \frac{dy}{dx} = \sec x $$
Final Answer:
$$ \frac{dy}{dx} = \sec x $$Problem 5 Solution (Image: 5.png)
Find $\frac{dy}{dx}$ for $y = e^{x \log a} + e^{a \log x} + e^{a \log a}$.
Solution Steps:
Simplify the terms using the property $e^{b \log c} = c^b$:
$ e^{x \log a} = a^x $
$ e^{a \log x} = x^a $
$ e^{a \log a} = a^a $ (This is a constant)
So, $y = a^x + x^a + a^a$.
Now, differentiate with respect to $x$:
$$ \frac{dy}{dx} = \frac{d}{dx}(a^x) + \frac{d}{dx}(x^a) + \frac{d}{dx}(a^a) $$
$$ \frac{d}{dx}(a^x) = a^x \log a $$
$$ \frac{d}{dx}(x^a) = a x^{a-1} $$
$$ \frac{d}{dx}(a^a) = 0 \quad (\text{since } a^a \text{ is a constant}) $$
Therefore,
$$ \frac{dy}{dx} = a^x \log a + a x^{a-1} $$
Final Answer:
$$ \frac{dy}{dx} = a^x \log a + a x^{a-1} $$Problem 6 Solution (Image: 6.png)
Find $\frac{dy}{dx}$ for $y = \frac{a^{2x} - a^{-2x}}{a^{2x} + a^{-2x}}$.
Solution Steps:
Let $v = 2x \log a$. Then $a^{2x} = e^v$ and $a^{-2x} = e^{-v}$. So $y = \frac{e^v - e^{-v}}{e^v + e^{-v}} = \tanh v$. $$ \frac{dy}{dv} = \text{sech}^2 v $$ $$ \frac{dv}{dx} = \frac{d}{dx}(2x \log a) = 2 \log a $$ Using the chain rule $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$: $$ \frac{dy}{dx} = \text{sech}^2 v \cdot (2 \log a) $$ Since $\text{sech}^2 v = 1 - \tanh^2 v = 1 - y^2$: $$ 1 - y^2 = 1 - \left(\frac{a^{2x} - a^{-2x}}{a^{2x} + a^{-2x}}\right)^2 = \frac{(a^{2x} + a^{-2x})^2 - (a^{2x} - a^{-2x})^2}{(a^{2x} + a^{-2x})^2} $$ Using $(A+B)^2 - (A-B)^2 = 4AB$: Let $A = a^{2x}$ and $B = a^{-2x}$. Then $AB = 1$. $$ 1 - y^2 = \frac{4}{(a^{2x} + a^{-2x})^2} $$ Therefore, $$ \frac{dy}{dx} = \frac{4}{(a^{2x} + a^{-2x})^2} (2 \log a) = \frac{8 \log a}{(a^{2x} + a^{-2x})^2} $$
Final Answer:
$$ \frac{dy}{dx} = \frac{8 \log a}{(a^{2x} + a^{-2x})^2} $$Problem 7 Solution (Image: 10.png)
Find $\frac{dy}{dx}$ for $y = (x \log x)^{\log(\log x)}$.
Solution Steps:
This is of the form $y = u^v$, so we use logarithmic differentiation. $$ \log y = \log ((x \log x)^{\log(\log x)}) $$ $$ \log y = \log(\log x) \cdot \log(x \log x) $$ $$ \log y = \log(\log x) \cdot (\log x + \log(\log x)) $$ Let $L = \log x$. Then $\log(\log x) = \log L$. $$ \log y = \log L \cdot (L + \log L) = L \log L + (\log L)^2 $$ Differentiate both sides with respect to $x$: $$ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (L \log L + (\log L)^2) $$ Use chain rule: $\frac{d}{dx} f(L) = \frac{d}{dL} f(L) \cdot \frac{dL}{dx}$, and $\frac{dL}{dx} = \frac{1}{x}$. $$ \frac{d}{dx}(L \log L) = (\log L + 1) \frac{1}{x} = (\log(\log x) + 1) \frac{1}{x} $$ $$ \frac{d}{dx}((\log L)^2) = (2 \log L \cdot \frac{1}{L}) \frac{1}{x} = \left(2 \frac{\log(\log x)}{\log x}\right) \frac{1}{x} $$ So, $$ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x}(\log(\log x) + 1) + \frac{1}{x} \frac{2 \log(\log x)}{\log x} $$ $$ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x \log x} [\log x (\log(\log x) + 1) + 2 \log(\log x)] $$ $$ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x \log x} [\log(\log x)(\log x + 2) + \log x] $$ $$ \frac{dy}{dx} = y \left( \frac{\log(\log x)(\log x + 2) + \log x}{x \log x} \right) $$ Substitute $y = (x \log x)^{\log(\log x)}$: $$ \frac{dy}{dx} = (x \log x)^{\log(\log x)} \left( \frac{(\log x + 2)\log(\log x) + \log x}{x \log x} \right) $$
Final Answer:
$$ \frac{dy}{dx} = (x \log x)^{\log(\log x)} \left( \frac{(\log x + 2)\log(\log x) + \log x}{x \log x} \right) $$Problem 8 Solution (Image: 12.png)
If $y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \dots \text{to } \infty}}}$, show that $(2y-1)\frac{dy}{dx} = \frac{1}{x}$.
Solution Steps:
The given expression for $y$ can be written recursively as: $$ y = \sqrt{\log x + y} $$ Assuming $y > 0$. Square both sides: $$ y^2 = \log x + y $$ Now, differentiate both sides with respect to $x$: $$ \frac{d}{dx}(y^2) = \frac{d}{dx}(\log x) + \frac{d}{dx}(y) $$ Using the chain rule for $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$: $$ 2y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} $$ Rearrange the terms to group $\frac{dy}{dx}$: $$ 2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} $$ $$ (2y - 1) \frac{dy}{dx} = \frac{1}{x} $$
Result to Show:
$$ (2y - 1) \frac{dy}{dx} = \frac{1}{x} $$Problem 9 Solution (Image: 13.png)
Find $\frac{dy}{dx}$ for $x = a \cos^3 t$, $y = a \sin^3 t$.
Solution Steps:
This is a parametric differentiation problem: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Find $\frac{dx}{dt}$:
$$ \frac{dx}{dt} = \frac{d}{dt}(a \cos^3 t) = a \cdot 3 \cos^2 t \cdot (-\sin t) = -3a \cos^2 t \sin t $$
Find $\frac{dy}{dt}$:
$$ \frac{dy}{dt} = \frac{d}{dt}(a \sin^3 t) = a \cdot 3 \sin^2 t \cdot (\cos t) = 3a \sin^2 t \cos t $$
Compute $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} $$
Assuming $\sin t \neq 0$ and $\cos t \neq 0$:
$$ \frac{dy}{dx} = \frac{\sin t}{-\cos t} = -\tan t $$
Final Answer:
$$ \frac{dy}{dx} = -\tan t $$Problem 10 Solution (Image: 14.png, first)
Find $\frac{dy}{dx}$ for $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$.
Solution Steps:
Parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
Find $\frac{dx}{d\theta}$:
$$ \frac{dx}{d\theta} = \frac{d}{d\theta}(a(\theta - \sin\theta)) = a(1 - \cos\theta) $$
Find $\frac{dy}{d\theta}$:
$$ \frac{dy}{d\theta} = \frac{d}{d\theta}(a(1 - \cos\theta)) = a(0 - (-\sin\theta)) = a \sin\theta $$
Compute $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = \frac{a \sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta} $$
Using half-angle identities $\sin\theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos\theta = 2 \sin^2(\theta/2)$:
$$ \frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)} $$
Assuming $\sin(\theta/2) \neq 0$:
$$ \frac{dy}{dx} = \frac{\cos(\theta/2)}{\sin(\theta/2)} = \cot(\theta/2) $$
Final Answer:
$$ \frac{dy}{dx} = \cot(\theta/2) $$Problem 11 Solution (Image: 14.png, second)
Find $\frac{dy}{dx}$ for $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$.
This problem is identical to Problem 10.
Final Answer:
$$ \frac{dy}{dx} = \cot(\theta/2) $$Problem 12 Solution (Image: 2.png, second, h=72)
Find $\frac{dy}{dx}$ for $x = \log t + \sin t$, $y = e^t + \cos t$.
Solution Steps:
Parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Find $\frac{dx}{dt}$:
$$ \frac{dx}{dt} = \frac{d}{dt}(\log t + \sin t) = \frac{1}{t} + \cos t $$
Find $\frac{dy}{dt}$:
$$ \frac{dy}{dt} = \frac{d}{dt}(e^t + \cos t) = e^t - \sin t $$
Compute $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = \frac{e^t - \sin t}{\frac{1}{t} + \cos t} = \frac{t(e^t - \sin t)}{1 + t \cos t} $$
Final Answer:
$$ \frac{dy}{dx} = \frac{t(e^t - \sin t)}{1 + t \cos t} $$Problem 13 Solution (Image: 2.png, third, h=73)
Find $\frac{dy}{dx}$ for $x = a (\cos t + \log \tan(t/2))$, $y = a \sin t$.
Solution Steps:
Parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Find $\frac{dy}{dt}$:
$$ \frac{dy}{dt} = \frac{d}{dt}(a \sin t) = a \cos t $$
Find $\frac{dx}{dt}$:
$$ \frac{dx}{dt} = \frac{d}{dt}(a (\cos t + \log \tan(t/2))) = a \left( -\sin t + \frac{1}{\sin t} \right) $$
(since $\frac{d}{dt}(\log \tan(t/2)) = \frac{1}{\sin t}$, as shown in Problem 2 derivation for $\log(\tan(x/2))$)
$$ \frac{dx}{dt} = a \left( \frac{-\sin^2 t + 1}{\sin t} \right) = a \frac{\cos^2 t}{\sin t} $$
Compute $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} $$
Assuming $\cos t \neq 0$:
$$ \frac{dy}{dx} = \frac{\cos t \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t $$
Final Answer:
$$ \frac{dy}{dx} = \tan t $$Problem 14 Solution (Image: 2.png, fourth, h=49)
Find $\frac{dy}{dx}$ for $x = \sin^{-1}\left(\frac{2t}{1+t^2}\right)$, $y = \tan^{-1}\left(\frac{2t}{1-t^2}\right)$.
Solution Steps:
Let $t = \tan\theta$. Assume $|t| < 1$, which implies $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.
For $x$:
$$ \frac{2t}{1+t^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta) $$
So $x = \sin^{-1}(\sin(2\theta)) = 2\theta = 2\tan^{-1}t$ (for $2\theta \in [-\pi/2, \pi/2]$).
$$ \frac{dx}{dt} = \frac{2}{1+t^2} $$
For $y$:
$$ \frac{2t}{1-t^2} = \frac{2\tan\theta}{1-\tan^2\theta} = \tan(2\theta) $$
So $y = \tan^{-1}(\tan(2\theta)) = 2\theta = 2\tan^{-1}t$ (for $2\theta \in (-\pi/2, \pi/2)$).
$$ \frac{dy}{dt} = \frac{2}{1+t^2} $$
For $|t|<1$:
$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2/(1+t^2)}{2/(1+t^2)} = 1 $$
Final Answer (for $|t|<1$):
$$ \frac{dy}{dx} = 1 $$Problem 15 Solution (Image: 2.png, fifth)
Differentiate $u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $v = \tan^{-1}x$. Find $\frac{du}{dv}$.
Solution Steps:
Let $x = \tan\theta$. Then $\theta = \tan^{-1}x = v$. Assume $x \neq 0$.
$\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = |\sec\theta|$. For $\theta \in (-\pi/2, \pi/2)$, $\sec\theta > 0$, so $|\sec\theta| = \sec\theta$.
The argument of $\tan^{-1}$ in $u$ becomes:
$$ \frac{\sec\theta - 1}{\tan\theta} = \frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}} = \frac{1 - \cos\theta}{\sin\theta} $$
Using half-angle identities:
$$ \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan(\theta/2) $$
So, $u = \tan^{-1}(\tan(\theta/2))$. Since $\theta \in (-\pi/2, \pi/2)$, $\theta/2 \in (-\pi/4, \pi/4)$.
Thus, $u = \theta/2$.
We have $u = \frac{\theta}{2}$ and $v = \theta$. So $u = \frac{1}{2}v$.
Differentiate $u$ with respect to $v$:
$$ \frac{du}{dv} = \frac{d}{dv}\left(\frac{1}{2}v\right) = \frac{1}{2} $$