How do you differentiate #y=(sinx)^(x)#?
Remembering the exponentiallogarithmic formula:
than our functio becomes:
and so, for the chain rule:
or, if you want:
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To differentiate ( y = (\sin x)^x ), you can use the chain rule along with the exponential differentiation rule.

First, take the natural logarithm of both sides to simplify the expression: [ \ln y = x \ln (\sin x) ]

Now, differentiate both sides with respect to ( x ): [ \frac{1}{y} \frac{dy}{dx} = \ln (\sin x) + x \frac{1}{\sin x} \cos x ]

Substitute back ( y = (\sin x)^x ) and solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} = (\sin x)^x \left( \ln (\sin x) + \frac{x \cos x}{\sin x} \right) ]
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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