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Y=ƒ(2x-1)/(x^2+1)and ƒ'(x)= sinx^2 then dy/dx Note x^2 means x square
I guess you mean dY/dx? If f'(x)= sinx^2 then dy/dx= sinx^2 => dy=sinx^2 dx , integrate to get the function y=f(x).Now calculate f(2x-1) and then do dY/dx.
I guess you mean dY/dx?
If f'(x)= sinx^2 then dy/dx= sinx^2 => dy=sinx^2 dx , integrate to get the function y=f(x).Now calculate f(2x-1) and then do dY/dx.
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