RD Sharma Solutions Class 12 Mathematics Chapter 10 VSA
RD Sharma Solutions Class 12 Mathematics Chapter 10 VSA
Updated on 20 Jan 2022, 06:07 PM IST
RD Sharma Class 12th Exercise VSA is the mostly used solutions material by the class 12 students. This set of books are an asset to any student who is preparing for the public exams. The concepts in mathematics chapter differentiation are well-covered in this book. This chapter consists of 14 exercises, ex 10.1 to 10.14. The VSA or the Very Short Answer part has 30 questions for which the answers can be found at the RD Sharma Class 12th Exercise VSA reference book.
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RD Sharma Class 12 Solutions Chapter 10 VSA Differentiation - Other Exercise
The solutions are provided by various experts in the domain of mathematics, this makes the RD Sharma Class 12th Exercise VSA Solutions book very much reliable. The answers are solved in various methods for the welfare of the students so that, they can rely on the ones that they feel easy.
RD Sharma Class 12 Solutions Chapter 10 VSA Differentiation - Other Exercise
Answer: The answer of given question is 1. Hint: $(f o f)(x)=f[f(x)]$ Given: $\text { If } f(x)=x+1 \text { then find } \frac{d}{d x}(\text { fof })(x)$ Solution: $\begin{aligned} &f(x)=x+1 \\\\ &f[f(x)]=f(x+1) \\\\ &=x+1+1 \end{aligned}$ $\begin{aligned} &\frac{d}{d x}(f o f)(x)=\frac{d}{d x}(x+2) \\\\ &=1 \end{aligned}$ so, the answer will be 1
Answer: The answer of the given question will be 1. Given: If $\sin ^{-1}(\sin x),-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ then write the value of $\frac{d y}{d x} \text { for } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Hint: $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Solution: $\begin{aligned} &\sin ^{-1}(\sin x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(x)=1 \\\\ &\therefore \frac{d y}{d x}=1 \end{aligned}$ So the answer will be 1.
Answer: The answer of the given question will be -1. Given: $\text { If } \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \text { and } y=\sin ^{-1}(\sin x) \text { find } \frac{d y}{d x}$ Hint: Here, $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ Solution: $\begin{aligned} &y=\sin ^{-1}(\sin x)=(\pi-x) \\\\ &\frac{d y}{d x}=\frac{d}{d x}(\pi-x)=-1 \end{aligned}$ So the answer of $\frac{d y}{d x}=-1$
Answer: The answer of the given question will be -1. Given: $\text { If } \pi \leq x \leq 2 \pi \text { and } y=\cos ^{-1}(\cos x) \text { find } \frac{d y}{d x}$ Hint: $\pi \leq x \leq 2 \pi$ Solution: $\begin{aligned} &y=\cos ^{-1}(\cos x)=(2 \pi-x) \\\\ &\frac{d y}{d x}=\frac{d}{d x}(2 \pi-x) \\\\ &=-1 \end{aligned}$ So the answer of $\frac{d y}{d x}=-1$
Answer: The answer of the given question will be -2x. Given: $\text { If } y=x|x| \text { find } \frac{d y}{d x} \text { for } x<0$ Hint: $|x|=\left\{\begin{array}{cc} x, & x \geq 0 \\ -x, & x<0 \end{array}\right\}$ Solution: $\begin{aligned} &y=x|x| \\\\ &y=x(-x,) \text { since } x<0 \\\\ &=-x^{2} \end{aligned}$ $\begin{aligned} &\frac{d y}{d x}=-\frac{d}{d x}\left(x^{2}\right)=-2 x \\\\ &\text { So, the } \frac{d y}{d x}=-2 x \end{aligned}$
Answer: The value of $\frac{d y}{d x} \text { at } x=e \text { is } 2 e^{e}$ Hint: Taking logarithm on both sides. Given: $y=x^{x}$ Solution: $y=x^{x}$ Taking logarithm on both sides. $\log y=x \log x$ Differentiating with respect to x on both sides., $\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{1}{x}+1 . \log x$ $\begin{aligned} &=1+\log x \\\\ &\Rightarrow \frac{d y}{d x}=y(1+\log x) \\\\ &\Rightarrow x^{x}(1+\log x) \end{aligned}$ So at $x=e$, $\begin{aligned} &\frac{d y}{d x}=e^{e}(1+\log e) \\\\ &=e^{e}(1+1) \\\\ &=2 e^{e} \end{aligned}$
Answer: $\frac{d y}{d x}=0, \text { for all } x \in R$ Hint: $-1<\frac{1-x^{2}}{1+x^{2}} \leq 1 \text { holds for all } x \in R$ Given: $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ Solution: $-1<\frac{1-x^{2}}{1+x^{2}} \leq 1 \text { holds for all } x \in R$ So, $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\frac{\pi}{2} ; \text { for all } x \in R$ $\therefore \sin ^{-1} m+\cos ^{-1} m=\frac{\pi}{2}, m \in(-1,1)$ Hence, $\frac{d y}{d x}=0, \text { for all } x \in R$
Answer: The value of $\frac{d y}{d x}=\frac{1}{x}, x \neq 0$ Hint: By using the properties of logarithm. Given: $y=\log |3 x|, x \neq 0$ Solution: $y=\log |3 x|$ So, $\begin{aligned} &\frac{d y}{d x}=\frac{1}{3 x} \cdot 3 \\\\ &=\frac{1}{x}, x \neq 0 \\\\ &\frac{d y}{d x}=\frac{1}{x}, x \neq 0 \end{aligned}$
Answer: $f^{\prime}(x)$is an odd function. Hint: Differentiate equation on the both sides with respect to x. Given: $f^{}(x)$is an even function. Solution: $f^{}(x)$is an even function. This means that $f(-x)=f(x)$ If we differentiate this equation on both sides with respect to x, we get. $\begin{aligned} &f^{\prime}(x) \cdot(-1)=f^{\prime}(x) \\\\ &\text { or },-f^{\prime}(-x)=f^{\prime}(x) \end{aligned}$ i.e. $f^{\prime}(x)$is an odd function.
Answer: $f^{\prime}(x)$is an even function Hint: Differentiate equation on the both sides with respect to x. Given: $f(x)$is an odd function. Solution: $f(x)$is an odd function. This means that $f(-x)=-f(x)$ If we differentiate this equation on both sides with respect to x, we get. $\begin{aligned} &f^{\prime}(-x) \cdot(-1)=-f^{\prime}(x) \\\\ &\text { or, } f^{\prime}(-x)=f^{\prime}(x) \end{aligned}$ i.e,$f^{\prime}(x)$is an even function
Answer: Derivation of $sin \; x$ with respect to $cos \; x$ is $-cot \; x.$ Hint: By using the chain rule of differentiation. Given: Derivation of $sin \; x$ with respect to $cos \; x$ is $\cos \; x.$ Solution: We have to find $\frac{d}{d(\cos x)}(\sin x)$ So, we use the chain rule of differentiation to evaluate this. $\frac{d}{d(\cos x)}(\sin x)=\frac{d(\sin x)}{d x} \cdot \frac{d x}{d(\cos x)}$ $\begin{aligned} &=\cos x \cdot \frac{1}{-\sin x} \\\\ &=-\cot x \end{aligned}$
Answer: The value of $\frac{d y}{d x} \text { is }-e^{x} \tan e^{x}$ Hint: Differentiating both sides with respect to x. Given: $y=\log \cos e^{x}$ Solution: $y=\log \cos e^{x}$ Differentiating both sides with respect to x. $\frac{d y}{d x}=\frac{d\left\{\log \left(\cos e^{x}\right)\right\}}{d x}$ $\frac{d y}{d x}=\frac{1}{\cos e^{x}} \cdot \frac{d\left(\cos e^{x}\right)}{d e^{x}} \cdot \frac{d e^{x}}{d x}\left(\therefore \frac{d \log x}{d x}=\frac{1}{x}\right)$ $=\frac{1}{\cos e^{x}} \cdot\left(-\sin e^{x}\right) \cdot e^{x}\left(\therefore \frac{d e^{x}}{d x}=e^{x}\right)$ $\begin{aligned} &=\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \\\\ &=-\tan e^{x} \cdot e^{x} \\\\ &=-e^{x} \cdot \tan e^{x} \end{aligned}$
This part consists of concepts that are present in every exercise of the chapter 10. Topics like differentiation of a function, inverse trigonometric functions, differentiation of infinite series, differentiation of parametric functions and many more are included in this section. Even though there is too much to learn, the RD Sharma Class 12 VSA material will require only less efforts from the students’ side.
Advantages that a student can gain while utilising the RD Sharma class 12 solutions:
Accurate answers:
The students need not worry regarding the accuracy of the solutions given in the RD Sharma class 12th exercise VSA books. All the answers are provided by a group of experts and professionals. Therefore, the steps in the RD Sharma class 12 Solutions Differentiation VSA are clearly explained in a way that the students can grasp easily.
Studying the concepts in advance:
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Not every concept in mathematics for the class 12 students requires to be taught be the teacher, there are easy topics that the students can learn on their own. This gives them the confidence to face such questions in the examination. These sums can be learnt before-hand using the Class 12 RD Sharma chapter 10 exercise VSA.
Practice sums:
Apart from the solved sums given for each question present in the book, the RD Sharma class 12th exercise VSA also contains extra sums for the students to test their understanding capability.
Multiple methods of solving a sum:
In mathematics, there are may methods in which a sum can be solved. The RD Sharma class 12th exercise VSA gives the exposure to the students of these various methods.
Improvement in scores:
Once the students use the RD Sharma books to practice the sums every day, they can gradually see the growth of their performance. Even the previous batch students have given positive feedback about this set of books.
Costless:
Who would not wish to have something for free of cost? The entire RD Sharma set of books can be found at the Career 360 website for free of cost. This is another strong reason why a student must make use of this wonderful opportunity.