# How do you take the derivative of #tan(4x)^tan(5x)#?

Let,

Taking natural log. ,both sides

#1/y(dy)/(dx)=4tan(5x)cos(4x)/sin(4x)xx1/cos^2(4x)+5sec^2(5x)lnt an(4x)#

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To find the derivative of ( \tan(4x)^{\tan(5x)} ), you would use the chain rule and the properties of logarithmic differentiation. First, rewrite the function using logarithms:

[ y = \tan(4x)^{\tan(5x)} ] [ \ln(y) = \ln(\tan(4x)^{\tan(5x)}) ] [ \ln(y) = \tan(5x) \cdot \ln(\tan(4x)) ]

Now, take the derivative of both sides with respect to ( x ):

[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[\tan(5x) \cdot \ln(\tan(4x))] ]

Apply the product rule on the right side:

[ \frac{1}{y} \cdot \frac{dy}{dx} = \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{d}{dx}[\ln(\tan(4x))] ]

Now, find the derivative of ( \ln(\tan(4x)) ) using the chain rule:

[ \frac{1}{y} \cdot \frac{dy}{dx} = \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 ]

Finally, solve for ( \frac{dy}{dx} ) by multiplying both sides by ( y ):

[ \frac{dy}{dx} = y \left( \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 \right) ]

[ \frac{dy}{dx} = \tan(4x)^{\tan(5x)} \left( \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 \right) ]

That's the derivative of ( \tan(4x)^{\tan(5x)} ).

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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.

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.

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.

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