Derivatives: Past years question for entrance exam(Online education)

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   Hi friends,in this blog I provide past years question of Mht CET of chapter derivatives  with proper solution in the from of PDF. The link of pdf given below in the next paragraph.this thing is very helpful for online education.

*Past years question of chapter Derivatives pdf link given below 👇👇 

                Download PDF derivatives

     

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1)mathematical logic

2)continuity

3)vectors

4)solid State

5)trigonometry functions

Properties of the derivative

We now consider various properties of differentiation. As we proceed, we will be able to differentiate wider and wider classes of functions.

Throughout this section and the next, we will be manipulating limits as we compute derivatives. We therefore recall some basic rules for limits; see the module Limits and continuity for details. The following hold provided the limits limx→af(x)$\underset{x\to a}{lim}f(x)$ and limx→ag(x)$\underset{x\to a}{lim}g(x)$ exist.

• The limit of a sum (or difference) is the sum (or difference) of the limits:
  limx→a(f(x)±g(x))=limx→af(x)±limx→ag(x).$$\underset{x\to a}{lim}(f(x)\pm g(x))=\underset{x\to a}{lim}f(x)\pm \underset{x\to a}{lim}g(x).$$
• The limit of a product is the product of the limits:
  limx→af(x)g(x)=(limx→af(x))(limx→ag(x)).$$\underset{x\to a}{lim}f(x)g(x)=(\underset{x\to a}{lim}f(x))(\underset{x\to a}{lim}g(x)).$$
  This includes the case of multiplication by a constant c$c$:
  limx→acf(x)=climx→af(x).$$\underset{x\to a}{lim}cf(x)=c\underset{x\to a}{lim}f(x).$$

The derivative of a constant multiple

Suppose we want to differentiate 4x7$4x^7$. Rather than returning to the definition of a derivative, we can use the following theorem.

Theorem

Let f$f$ be a differentiable function and let c$c$ be a constant. Then the derivative of cf(x)$cf(x)$ is cf′(x)$c{f}^{\prime }(x)$. That is,

ddx[cf(x)]=cf′(x).$${\displaystyle \frac{d}{dx}}[cf(x)]=c{f}^{\prime }(x).$$

This fact can also be written in Leibniz notation as

ddx(cf)=cdfdx.$${\displaystyle \frac{d}{dx}}(cf)=c{\displaystyle \frac{df}{dx}}.$$

Proof

The derivative of cf(x)$cf(x)$ is given by

limΔx→0cf(x+Δx)−cf(x)Δx=climΔx→0f(x+Δx)−f(x)Δx=cf′(x).$$\underset{\mathrm{\Delta }x\to 0}{lim}{\displaystyle \frac{cf(x+\mathrm{\Delta }x)-cf(x)}{\mathrm{\Delta }x}}=c\underset{\mathrm{\Delta }x\to 0}{lim}{\displaystyle \frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}=c{f}^{\prime }(x).$$

We may factor out the c$c$, since it is just a constant.

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Example

What is the derivative of 4x7$4x^7$?

Solution

The theorem tells us that the derivative of 4x7$4x^7$ is 4 times the derivative of x7$x^7$. Hence,

ddx(4x7)=4ddx(x7)=4⋅7x6=28x6.$$\begin{array}{rl}{\displaystyle \frac{d}{dx}}(4x^7)& =4{\displaystyle \frac{d}{dx}}(x^7)\\ & =4\cdot 7x^6\\ & =28x^6.\end{array}$$

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