How do you differentiate #f(x) = ln(sqrt(arcsin(e^(2-x^2)) ) # using the chain rule?

Answer 1

#f'(x)=(-xe^2)/(arcsin(e^(2-x^2))sqrt(e^(2x^2)-e^4))#

First, recognize that this can be written as

#f(x)=1/2ln(arcsin(e^(2-x^2)))#
since the square root is really a power of #1"/"2#, and powers can be brought out of logarithms.
The first issue is the natural logarithm. But, according to the chain rule, #d/dx[ln(u)]=(u')/u#. Thus,
#f'(x)=1/2(d/dx[arcsin(e^(2-x^2))])/arcsin(e^(2-x^2))#
The next issue lies in differentiating the arcsine function. The following rule can be used: #d/dx[arcsin(u)]=(u')/sqrt(1-u^2)#
#f'(x)=1/(2arcsin(e^(2-x^2)))*(d/dx[e^(2-x^2)])/sqrt(1-(e^(2-x^2))^2)#
To differentiate the #e# power, use the chain rule once more: #d/dx[e^u]=e^u*u'#
#f'(x)=1/(2arcsin(e^(2-x^2)))*(d/dx[2-x^2]e^(2-x^2))/sqrt(1-e^(4-2x^2))#
#f'(x)=1/(2arcsin(e^(2-x^2)))*(-2xe^(2-x^2))/sqrt(1-e^(4-2x^2))#

Nasty simplification from here on out, now that the calculus is over...

#f'(x)=(-xe^(2-x^2))/(arcsin(e^(2-x^2))sqrt(1-e^(4-2x^2))#
#f'(x)=((-xe^2)/(e^(x^2)))/(arcsin(e^(2-x^2))sqrt(1-e^4/e^(2x^2)))#
#f'(x)=((-xe^2)/(e^(x^2)))/(arcsin(e^(2-x^2))sqrt((e^(2x^2)-e^4)/e^(2x^2)))#
#f'(x)=((-xe^2)/(e^(x^2)))/(arcsin(e^(2-x^2))(1/e^(x^2))sqrt(e^(2x^2)-e^4))#
#f'(x)=(-xe^2)/(arcsin(e^(2-x^2))sqrt(e^(2x^2)-e^4))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To differentiate ( f(x) = \ln(\sqrt{\arcsin(e^{2-x^2})}) ) using the chain rule, follow these steps:

  1. Identify the outer function as ( \ln(u) ) and the inner function as ( \sqrt{\arcsin(v)} ).
  2. Compute the derivative of the outer function with respect to its argument ( u ), which is ( \frac{1}{u} ).
  3. Compute the derivative of the inner function with respect to its argument ( v ), which is ( \frac{1}{2\sqrt{v(1-v)}} ).
  4. Compute the derivative of the innermost function with respect to ( x ) using the chain rule.
  5. Substitute the derivatives into the chain rule formula: ( f'(x) = \frac{1}{\sqrt{\arcsin(e^{2-x^2})}} \cdot \frac{1}{2\sqrt{\arcsin(e^{2-x^2})\left(1-\arcsin(e^{2-x^2})\right)}} \cdot \frac{d}{dx}\left(\arcsin(e^{2-x^2})\right) ).

Then, compute ( \frac{d}{dx}\left(\arcsin(e^{2-x^2})\right) ) using the chain rule and simplify the expression.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7