Find the gradient of the curve y=2sinx/x^3 at the point x=
- To find the gradient of the curve we must differentiate the function. The function is in the form of a quotient, y=u/v, where u and v are functions of x. Therefore, we can use the quotient rule, dy/dx = (v (du/dx) – u (dv/dx))/ v^2. 2) We can write u = 2sinx and differentiating this we obtain du/dx = 2cosx. 3) We then take v= x^3 and differentiating this we obtain dv/dx = 3x^2 by multiplying by the power then taking one off the power (the general rule for differentiation being y=ax^n, dy/dx = anx^(n-1). 4) The quotient rule then gives, dy/dx = (v (du/dx) – u (dv/dx))/ v^2 = ( 2x^3cosx – 6x^2sinx) / x^6 = 2x^2 (xcosx – 3sinx) / x^6 = 2(xcosx – 3sinx)/x^4. 5) To find the gradient at the point Q where x=1 we substitute x=1 into dy/dx. We obtain, dy/dx = 2(cos1 – 3sin1).
HM
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