Question

Differentiate ( sin2x )^x + sin^-1 √3x​

Answer 1

Answer:

plzzz give me brainliest ans and plzzzz follow me

Answer 2

Answer: The final expression for the derivative is:

2xcos(2x) (sin2x)^(x-1) + 3/(2√(1-3x))

To differentiate the expression, we will need to use the chain rule and the product rule. Let's start with the first term:

( sin2x )^x

Using the chain rule, we have:

= x( sin2x )^(x-1) * d/dx (sin2x)

Now, applying the chain rule to the derivative of sin2x, we have:

= x( sin2x )^(x-1) * cos(2x) * d/dx (2x)

= 2xcos(2x) (sin2x)^(x-1)

Next, let's look at the second term:

(sin^-1)√3x

Using the chain rule, we have:

= d/dx (sin^-1) √3x * d/dx (√3x)

= 1/√(1-(√3x)^2) * 3/2x^(1/2)

= 3/(2√(1-3x))

Putting the two terms together, the final expression for the derivative is:

2xcos(2x) (sin2x)^(x-1) + 3/(2√(1-3x))

Learn more about chain rule here

https://brainly.in/question/48504877

Learn more about derivatives here

https://brainly.in/question/758525

#SPJ2