Differentiate √(30+sin²x) with respect to x

Answer 1

To differentiate √(30+sin²x) with respect to x, we can use the chain rule.

Let's denote the function as f(x) = √(30+sin²x).

To differentiate f(x) with respect to x, we differentiate the outer function (square root) and then multiply by the derivative of the inner function (30+sin²x).

The derivative of the square root function is 1/2√(30+sin²x).

The derivative of the inner function, 30+sin²x, with respect to x is 2cosx * (-sinx) = -2sinxcosx.

Multiplying these two results together, we have:

f'(x) = -2sinxcosx * 1/2√(30+sin²x)

Simplifying further:

f'(x) = -sinx * cosx * 1/√(30+sin²x)

Therefore, the differentiated form of √(30+sin²x) with respect to x is -sinx * cosx / √(30+sin²x).

Answer 2

let p = (30+sin²x)
then we want
d/dx ( p^.5) = .5 p^-.5 dp/dx
but dp/dx = 2 sin x cos x
so
.5 (1/p^.5) 2 sin x cos x
= sin x cos x / √(30+sin²x)

Answer 3

Apologies for the mistake in my previous response. You are correct.

Let p = (30+sin²x).

We want to find d/dx (p^.5).

Using the chain rule, we have:

d/dx (p^.5) = .5 p^(-.5) dp/dx

Now, let's find dp/dx.

dp/dx = d(30+sin²x)/dx = 2sinx*cosx

Substituting this back into the previous equation, we have:

.5 p^(-.5) dp/dx = .5 (1/√p) * 2sinx*cosx

Simplifying further:

= sinx*cosx / √(30+sin²x)

Therefore, the correct differentiated form of √(30+sin²x) with respect to x is sinx*cosx / √(30+sin²x).