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sika f((3)/(2x-3))=(2x+3)/(x+4). f^(')(1)?

sika f(32x−3)=2x+3x+4 f\left(\frac{3}{2 x-3}\right)=\frac{2 x+3}{x+4} .\newlinef′(1)? f^{\prime}(1) ?

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Q. sika f(32x−3)=2x+3x+4 f\left(\frac{3}{2 x-3}\right)=\frac{2 x+3}{x+4} .\newlinef′(1)? f^{\prime}(1) ?
  1. Find Derivative of f: First, let's find the derivative of the function ff with respect to xx.f(32x−3)=2x+3x+4f\left(\frac{3}{2x-3}\right)=\frac{2x+3}{x+4}To find f′(x)f'(x), we need to differentiate both sides with respect to xx.
  2. Apply Quotient Rule: Differentiate the right side using the quotient rule: (v′u−uv′)/v2(v'u - uv') / v^2 Let u=(2x+3)u = (2x+3) and v=(x+4)v = (x+4) u′=2u' = 2 and v′=1v' = 1
  3. Calculate f′(x)f'(x): Now apply the quotient rule:\newlinef′(32x−3)=[(x+4)(2)−(2x+3)(1)](x+4)2f'\left(\frac{3}{2x-3}\right) = \frac{[(x+4)(2) - (2x+3)(1)]}{(x+4)^2}\newline= 2x+8−2x−3(x+4)2\frac{2x+8 - 2x - 3}{(x+4)^2}\newline= 5(x+4)2\frac{5}{(x+4)^2}
  4. Find xx for f′(1)f'(1): We need to find f′(1)f'(1), but we have f′(32x−3)f'\left(\frac{3}{2x-3}\right). So we need to find the xx that makes 32x−3\frac{3}{2x-3} equal to 11.\newlineSet 32x−3=1\frac{3}{2x-3} = 1 and solve for xx.\newline2x−3=32x - 3 = 3\newlinef′(1)f'(1)00\newlinef′(1)f'(1)11

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