How do you differentiate #y=(sinx)^lnx#?

Answer 1

# dy/dx = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x} #

Generally when dealing with a variable exponent it is easier to differentiate (and understand) by taking natural logarithms (to remove the exponent) and differentiating implicitly:

We have:

# y = (sinx)^(lnx) #

Take Natural logarithms:

# lny = ln{(sinx)^(lnx)} # # \ \ \ \ \ \= (lnx)(ln(sinx)) \ \ \ # (rule of logs)
Differentiate wrt #x# (LHS implicitly; RHS product rule with chain rule):
# \ 1/ydy/dx = (lnx)(1/sinx*cosx) + (1/x)(ln(sinx)) # # \ \ \ \ \ \ \ \ \ \ \= (lnx)(cotx) + (ln(sinx))/x #
# :. dy/dx = y{(lnx)(cotx) + (ln(sinx))/x} # # \ \ \ \ \ \ \ \ \ \ = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x} #
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Answer 2

To differentiate ( y = (\sin x)^{\ln x} ), you would use the chain rule along with logarithmic differentiation. Here's the step-by-step process:

  1. Take the natural logarithm of both sides of the equation to simplify the function. [ \ln y = \ln \left( (\sin x)^{\ln x} \right) ]

  2. Apply the logarithm property (\ln(a^b) = b \ln(a)) to simplify the expression. [ \ln y = \ln x \cdot \ln(\sin x) ]

  3. Differentiate both sides of the equation with respect to (x). [ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} \cdot \ln(\sin x) + \frac{\ln x}{\sin x \cdot \cos x} ]

  4. Solve for ( \frac{dy}{dx} ). [ \frac{dy}{dx} = y \cdot \left( \frac{1}{x} \cdot \ln(\sin x) + \frac{\ln x}{\sin x \cdot \cos x} \right) ]

  5. Substitute back the original function (y = (\sin x)^{\ln x}) to get the final result. [ \frac{dy}{dx} = (\sin x)^{\ln x} \left( \frac{1}{x} \cdot \ln(\sin x) + \frac{\ln x}{\sin x \cdot \cos x} \right) ]

That's the derivative of (y) with respect to (x).

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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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