Test: Derivatives: Sum And Difference Rule - JEE MCQ

dy/dx = 40x^½ - 20
dy/dx = 40x^½-1 * 20 - 0
= 40/2(x)^½
20/(x)1/2

Since the number of terms is finite, just differentiate term by term using the power rule.
 dy/dx = 0 + 1 + 2x/2! + 3x2/3! + ... + nx^n-1/n!
dy/dx = 1 + x + x2/2! + ... + xn-1/(n -1)!
Compare dy/dx to y and note that the last term of y, which is xn/n!, is not in dy/dx, while all the other terms of y are in dy/dx.
Then dy/dx = y - xⁿ/n! 
⇒ dy/dx + xⁿ/n! = y

y = cosecx
= 1/sinx= (sinx)⁻¹
⇒ dy/dx=−(sinx)⁻² × d/dx(sinx)
= −cosx/sin^2x
= −cosx/sinx ×(1/sinx)
= −cot x cosec x

The derivative of the function 2sin(x) + x with respect to x is 2cos(x) + 1.

y = 5 sin x - 11 cosx + 1/x²
dy/dx = 5 cosx - 11(-sinx) + (-2/x³)
= 5cosx + 11sinx - 2/x³

dy/dx = 3x²-x
dy/dx = 3(2)x^(2-1) - 1x^(1-1)
= 6x-1

Y=sinx+cosx
dy/dx=cosx-sinx

2x² + 3x - 3 =0
dy/dx = 4x + 3 
At x = -2
4(-2) + 3 = -5

Y=x³-20x
dy/dx=3x²-20
(dy/dx) at x=10 is
=3×100-20
=280