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How do you differentiate #sqrt(sin^3(1/x) # using the chain rule?

Answer 1

Elijah Abram

the answer
#dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(1/x))]#

Firstly suppose: #u=1/x#

#(du)/dx=-1/x^2#

#r=sin^3(u)#

#(dr)/(du)=3sin^2(u)*cos(u)#

#y=sqrtr#

#dy/(dr)=1/(2sqrtr)#

now

#dy/(dx)=dy/(dr)*(dr)/(du)(du)/dx#

#dy/dx=1/(2sqrtr)*3sin^2(u)*cos(u)[-1/x^2]#

#dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(1/x))]#

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Answer 2

Elijah Abram

To differentiate ( \sqrt{\sin^3(1/x)} ) using the chain rule, follow these steps:

Let ( u = \sin(1/x) ).
Rewrite the expression as ( y = \sqrt{u^3} ).
Differentiate ( y ) with respect to ( u ): ( \frac{dy}{du} = \frac{3u^{\frac{3}{2}}}{2\sqrt{u}} ).
Differentiate ( u ) with respect to ( x ): ( \frac{du}{dx} = -\frac{\cos(1/x)}{x^2} ).
Apply the chain rule: ( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} ).
Substitute the expressions for ( \frac{dy}{du} ) and ( \frac{du}{dx} ) into the equation.
Simplify the result to obtain the final derivative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.