Differentiate arcsin(3x)?

Question

I'm trying to differentiate arcsin3x (with respect to x)
The answer is -24/sqrt(1-(3x)^2) but I can't get it right.

I tried the product rule and got 3x.1/sqrt(1+x^2)+3arcsin

Any chance of showing my the correct steps?

Answer 1

y = arcsin 3x

Because differentiating an inverse trigonometric function isnt so obvious.... and in fact, to prove the derivative of an inverse function....

Simplify to a more familiar, easily derived equation...
sin y = 3x

Taking the derivative of arcsin x with respect to x is no different than taking the derivative of sin y (or sin f(x) ) with respect to x.

∂/∂x sin y = ∂/∂x 3x

With basic and implicit differentiation...
y' ⋅ cos y = 3

Solving for y'...
y' = 3 ⁄ cos y

And plugging in for y the original function...
y' = 3 ⁄ cos (arcsin 3x)

You could leave it at that... or with properties of trigonometric functions (the cosine of an arcsin)...
y' = 3 ⁄ √[ 1 − (3x)² ]
y' = 3 ⁄ √[ 1 − 9x² ]


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The derivative of arcsin f(x)

∂/∂x arcsin f(x) = [ ∂/∂x f(x) ] ⁄ √[ 1 − f(x)² ]

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The book is wrong about the -24...

And Im not sure how you applied the product rule... "3x" and "arcsin" are not being multiplied together. "arcsin" is a trigonometric function, not a factor

Answer 2

Derivative Of Arcsin

Answer 3

f(x) = arcsin(3x)

f '(x) = d/dx arcsin(3x)

since the derivative of arcsinx = 1/(sqrt(1 - x^2))

so the derivative of arcsin3x would become 1/(sqrt(1 - (3x)^2))

Using Chain Rule

= 1/(sqrt(1 - (3x)^2)) * 3

= 3/(sqrt(1 - 9x^2))

Answer 4

When doing anc functions, you can proceed as follows:
y = arcsin(3x)
so:
sin y = 3x
cos y * dy/dx = 3
dy/dx = 3/ cosy
Now I need to express cos y in terms of x:
If the sin y = 3x
think of y as the angle, at distance 1 from the origin. with coordinates: (a,b)
b = 3x
by the pathagorean theorem a = √( 1 - 9x^2)
so cos y = √( 1 - 9x^2)
by substitution we have the derivative:
dy/dx = 3/√( 1 - 9x^2).
I have looked at my answer and am confident it is right I don't see how the auther got -24 in the numerator.
The formula is d(arc sin u)/dx = 1/√(1-u^2)*du/dx

Answer 5

no product rule necessary, only chain rule

basic form for arcsin(u) is u'/sqrt(1-u^2)

so for arcsin(3x) you will get 3/sqrt(1-9x^2)

don't know where the -24 is coming from...

Answer 6

That's not the right answer.

The derivative of sin^-1 x = 1/√(1 - x²)

For sin^-1 (3x), let u = 3x

Then d/dx sin^-1 (u) = [1/√(1 - u²)] * du/dx

= 3/√[1 - (3x)² ]

You can even work it from first principles:

Let y = sin^-1 (3x)

This means: sin y = 3x

d/dy sin y = cosy dy/dx = 3

dy/dx = 3/cos y

Now if sin y = 3x, make a right triangle with 1 as the hypotenuse and 3x as the opposite side. The sin of this triangle is 3x. The adjacent side is therefore (by Pythagoras) √(1-(3x)²)

So cos y = √(1-(3x)²)

and

dy/dx = 3/√(1-(3x)²)

Answer 7

arcsin(3x)

differentiating

1/sqrt[1 - (3x)^2] *d(3x)

=>3/sqrt(1 - 9x^2)

Answer 8

d/dx(arcsin3x) = 3/sqrt(1-9x^2)

-24/sqrt(1-(3x)^2) is incorrect

Answer 9

d/dx (arcsin 3x)
= 1 / √[ 1 - (3x)^2] * d/dx (3x)
= 3 / √( 1 - 9x^2)

Answer 10

how the heck they have that answer

correct answer will be 3/(Squareroot(1-9x^2))