Question

d/dx 2 sin^2x.....?​

Answer 1

Answer:

2Sin2x

Step-by-step explanation:

To find--->

------------

Differential coefficient of 2sin²x

Solution--->

-------------

d

----- (2Sin²x)

dx

d

= 2 ----- ( Sinx)²

dx

d/dx of( xⁿ)=n xⁿ⁻¹

d

= 2 × 2 (Sinx)²⁻¹ ------ (Sinx)

dx

= 4 Sinx Cosx

We have a formula 2SinACosA=Sin2A

=2(2Sinx Cosx)

=2 Sin2x

Additional information--->

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1)d/dx (Cosx)=-Sinx

2)d/dx (tanx)=Sec²x

3)d/dx (Secx)=Secx tanx

4)d/dx(Cotx)=-Cosec²x

5)d/dx(Cosecx)=-Cosecx Cotx

6)d/dx(Sin⁻¹x)=1/√(1-x²)

7)d/dx(Cos⁻¹x)=-1/√(1-x²)

8)d/dx(tan⁻¹x)=1/(1+x²)

9)d/dx(Cot⁻¹x)=-1/(1+x²)