How do you find the gradient of a line at a certain point when f(x) is in the form of a fraction, where both the numerator and denominator are functions of x?

Take the example f(x)=(2x^2+1)/(3x+5) , where we're finding the gradient at x=0. First, you need to differentiate f(x) to get f'(x). Because f(x) is a fraction where both the numerator and denominator are functions of x, we use the quotient rule. This gives us f'(x)=((3x+5)(4x)-(2x^2+1)(3))/(3x+5)^2 Now, we plug in the value of x, since f'(x) gives us the gradient. So f'(0)=((30+5)(40)-(20^2+1)(3))/(30+5)^2 f'(0)=-3/25 This means the gradient of f(x) at x=0 is -3/25