Differentiation and it's Application: ( Prepared by :NMK)

5x+2

 $2x^3+5$  

4x+5

5+4x

2x+2

2x+1

1+2x

2x+4

 $\frac{145}{23}$  

 $\frac{27}{133}$  

 $\frac{55}{27}$  

 $\frac{133}{27}$  

 $-\frac{9}{4}$  

 $-\frac{4}{9}$  

 $\frac{9}{4}$  

 $6x^2\left(2x^3+7\right)^5$  

 $36x^2\left(2x^2+7\right)^5$  

 $36x\left(2x^3+7\right)^{5^{ }}$  

 $36x^2\left(2x^3+7\right)^5$  

(2,3)

(2,2)

(2,1)

(3,2)

 $2\left(x+7\right)^9\left(x^2+2\right)^6\left(12x^2+49x+10\right)$  

 $2\left(x+7\right)^8\left(x^2+2\right)^5\left(12x^2+49x+10\right)$  

 $2\left(x+7\right)^{10}\left(x^2+2\right)^6\left(24x^2+49x+10\right)$  

 $6\left(x+7\right)^9\left(x^{ }+2\right)^6\left(12x^2+49x+10\right)$  

 $-\frac{23}{\left(3x-4\right)^3}$  

 $-\frac{32}{\left(3x-4\right)^2}$  

 $\frac{23}{\left(3x-4\right)^2}$  

 $-\frac{23}{\left(3x-4\right)^2}$  

 $\frac{\text{d}^2v}{\text{d}t^2}=100$  

 $\frac{\text{d}^2v}{\text{d}t^2}=-20$  

 $\frac{\text{d}^2v}{\text{d}t^2}=-10$  

 $\frac{\text{d}^2v}{\text{d}t^2}=10$  

4%

5%

-4%

6%