How do you differentiate #g(x) = sqrt(2x^2-1)cos3x# using the product rule?

Answer 1

#g' (x)=-3*sin (3x)sqrt(2x^2-1)+ (2x*cos 3x)/(sqrt(2x^2-1))#

The formula for the product rule is

#d/dx(uv)=u dv/dx+ v*du/dx#
from the given #g(x)=sqrt(2x^2-1)*cos 3x#
We let #u=sqrt(2x^2-1)# and #v=cos 3x#
#d/dx(uv)=u dv/dx+ v*du/dx#
#g' (x)=d/dx(sqrt(2x^2-1)*cos 3x)=(sqrt(2x^2-1)) d/dx(cos 3x)+ (cos 3x)*d/dxsqrt(2x^2-1)#
#g' (x)=# #(sqrt(2x^2-1))*(-3 sin 3x)+ (cos 3x)*1/(2*sqrt(2x^2-1))*(4x-0)#
#g' (x)=-3*sin (3x)sqrt(2x^2-1)+ (2x*cos 3x)/(sqrt(2x^2-1))#

God bless....I hope the explanation is useful.

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Answer 2

To differentiate g(x) = √(2x^2 - 1) * cos(3x) using the product rule:

  1. Let u(x) = √(2x^2 - 1) and v(x) = cos(3x).
  2. Find the derivatives of u(x) and v(x): u'(x) = (4x) / (2√(2x^2 - 1)) and v'(x) = -3sin(3x).
  3. Apply the product rule: g'(x) = u(x)v'(x) + v(x)u'(x).
  4. Substitute the values obtained in steps 2 and 3 into the product rule formula.

g'(x) = (√(2x^2 - 1))(-3sin(3x)) + (cos(3x))((4x) / (2√(2x^2 - 1)))

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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.

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