Question #190574

State the chain rule, the product rule and the quotient rule of differentiation. For each rule, give an example of a trigonometric expression which can be differentiated using the rule.


Expert's answer

Chain rule of differentiation:


dydx=dydududx\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}


Example: f(x)=sin(3x2+x)f(x) = sin(3x^2+x)


It looks like the outside function is the sine and the inside function is 3x2+x. The derivative is then.


f′(x)=cos(3x2+x)(6x+1)f'(x) = cos(3x^2+x)(6x+1)


=(6x+1)cos(3x2+x)= (6x+1)cos(3x^2+x)


Product Rule of Differentiation:


d(uv)dx=udvdx+vdudx\dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}


Example : f(x)=xsinxf(x) = xsinx


f′(x)=xd(sinx)dx+sinxd(x)dxf'(x) = x\dfrac{d(sinx)}{dx} + sinx \dfrac{d(x)}{dx}


f′(x)=xcosx+sinx.f'(x) = xcosx+sinx.


Quotient Rule of differentiation:


dydx=vdudx−udvdxv2\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}


Example: f(x)=sinxxf(x) = \dfrac{sinx}{x}


f′(x)=xd(sinx)dx−sinxd(x)dxx2f'(x) = \dfrac{x\dfrac{d(sinx)}{dx}-sinx \dfrac{d(x)}{dx}}{x^2}


f′(x)=xcosx−sinxx2f'(x) = \dfrac{xcosx-sinx}{x^2}


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