IB DP Math: AA SL :Topic SL 5.1-gradient of function: Study Notes IB DP PhysicsIB DP MathsIB DP ChemistryIB DP Biology IB DP Math SL: Study Notes- All Chapters IB DP Maths AA&AI: IB Style Question BankIB DP Physics SL&HL: IB Style Question BankIB DP Biology SL&HL: IB Style Question BankIB DP Chemistry SL&HL: IB Style Question Bank SL 5.1 Derivative interpreted as gradient function and as rate of change. ContentDerivative interpreted as gradient function and as rate of change.Understandings:Rates of ChangeRATE OF CHANGE (OR GRADIENT) IN A STRAIGHT LINERATE OF CHANGE (OR GRADIENT) ΙΝ Α CURVEAverage Rates of Change: Motionaverage velocityInstantaneous rate of changeTHE GRADIENT OF A TANGENTGuidance, clarification and syllabus linksForms of notation: \(\frac{dy}{dx},f'(x),\frac{dV}{dr}or \frac{ds}{dt}\) for the first derivative.Informal understanding of the gradient of a curve as a limit. Question[with GDC]Let \(f(x)=\frac{x^{3}+1}{\sin x}\)(a) Find \(f'(x)\).(b) Find the gradient of the curve \(y=f(x)\)(ⅰ) at \(x=\frac{\pi }{4}\) (ⅰⅰ) at \(x=1 rad\).Answer/ExplanationAns(a) \(f'(x)=\frac{3x^{2}\sin x-(x^{3}+1)\cos x}{\sin ^{2}x}\)(b) Directly by GDC (i) \(f'(\frac{\pi }{4})\cong 0.518\) (ii) \(f'(1)\cong 2.04\)[Notice: the exact value for (i) is \(f'(\frac{\pi }{4})=\frac{3\pi ^{2}}{16}\sqrt{2}-\frac{\pi ^{3}+64}{64}\sqrt{2}]\) More IB Style Math AA SL Questions..