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How do you find the derivative of #y = x^(cos x)#?

Answer 1

Luke Alvarez

#dy/dx=x^cosx((cosx)/x-lnxsinx)#

Take the natural log of both sides and move the cos in front of the natural log.

#lny=(cosx)lnx#

Use implicit differentiation. You have to take the product rule of the right side.

#(1/y)(dy/dx)=(cosx)(1/x)+(lnx)(-sinx)#

#(1/y)(dy/dx)=(cosx)/x-lnxsinx#

Solve for #dy/dx#

#dy/dx=y((cosx)/x-lnxsinx)#

Plug in #x^cosx# for y

#dy/dx=x^cosx((cosx)/x-lnxsinx)#

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Answer 2

Evelyn Agard

To find the derivative of ( y = x^{\cos(x)} ), you can use the chain rule. The chain rule states that if ( u(x) ) and ( v(x) ) are differentiable functions of ( x ), then the derivative of ( u(v(x)) ) with respect to ( x ) is ( u'(v(x)) \cdot v'(x) ). Applying the chain rule to ( y = x^{\cos(x)} ), we have:

( \frac{dy}{dx} = \frac{d}{dx} \left(x^{\cos(x)}\right) = \cos(x) \cdot x^{\cos(x)-1} - \ln(x) \cdot x^{\cos(x)} \cdot \sin(x) )

So, the derivative of ( y = x^{\cos(x)} ) with respect to ( x ) is ( \cos(x) \cdot x^{\cos(x)-1} - \ln(x) \cdot x^{\cos(x)} \cdot \sin(x) ).

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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.