30. If (sinx)y=ycosx, then find dxdy.
(differentiation, implicit differentiation, logarithmic differentiation, chain rule, trigonometric functions)
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Step-by-Step Solution
Step 1: Apply Natural Logarithm
The given equation involves variables in both the base and the exponent, which suggests using logarithmic differentiation. Taking the natural logarithm on both sides allows us to use the logarithm property ln(ab)=blna to bring down the exponents.
Step 2: Differentiate Implicitly
Now we differentiate both sides of the equation with respect to x. Since y is a function of x, we must use the product rule and chain rule for terms involving y.
Step 3: Apply Product Rule on Left Side
For the left side, we apply the product rule dxd(uv)=u′v+uv′ where u=y and v=ln(sinx). The derivative of ln(sinx) requires the chain rule: sinx1⋅cosx.
Step 4: Apply Product Rule on Right Side
For the right side, we also apply the product rule where u=cosx and v=lny. The derivative of cosx is −sinx. The derivative of lny with respect to x requires the chain rule: y1dxdy.
Step 5: Combine and Rearrange Terms
Now we equate the results from differentiating both sides. We need to isolate dxdy. To do this, we gather all terms containing dxdy on one side of the equation and move all other terms to the opposite side. We also simplify sinxcosx to cotx.
Step 6: Factor out dxdy
Factor out dxdy from the terms on the left side of the equation. This makes it easier to solve for dxdy.
Step 7: Solve for dxdy
Finally, divide both sides by the factor multiplied by dxdy to get the expression for dxdy. To simplify the complex fraction, we multiply the numerator and denominator by y to eliminate the fraction within the denominator.