Derivative of e^(-(1/3)x^2-(1/5)y^4) ?

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In summary, the derivative of e^(-(1/3)x^2-(1/5)y^4) is {-(2/3)x-[((4/5)y^3)dy/dx]}e^(-(1/3)x^2-(1/5)y^4).
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Derivative of e^(-(1/3)x^2-(1/5)y^4) ?

I need help figuring this out. I know that the derivative of e^5x is 5e^x but I'm not sure what to do when x is raised to a power or when there is another variable involved, such as y. Once again here is what I need to find: Derivative of e^(-(1/3)x^2-(1/5)y^4)

And if you guys know of a way that I can display that in a clearer way such as with some type of symbol tool please let me know for future posts. To make that clearer I want to find the derivative of e raised to the power of (negative one-third x^2 minus one-fifth y^4). Also, if you know of a site that fully explains the derivatives of e and includes explanations to the above problem let me know. Thanks.
 
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It may just be a typo on your part, but the derivative of e^5x is 5e^5x, not 5e^x.

With derivatives of e^(stuff), you start out by simply re-writing e^(stuff) and then multiplying by the derivative of (stuff). Like in e^5x, you re-write e^5x and then multiply by the derivative of 5x, which is five, giving you 5e^5x.

In your problem, you re-write your original term and then multiply by the derivative of
(-(1/3)x^2-(1/5)y^4) which is (-(2/3)x-[((4/5)y^3)dy/dx]) using implicit differentiation. The final answer would then be {-(2/3)x-[((4/5)y^3)dy/dx]}e^(-(1/3)x^2-(1/5)y^4).
 
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FAQ: Derivative of e^(-(1/3)x^2-(1/5)y^4) ?

What is the formula for the derivative of e^(-(1/3)x^2-(1/5)y^4)?

The formula for the derivative of e^(-(1/3)x^2-(1/5)y^4) is e^(-(1/3)x^2-(1/5)y^4) * (-2/3)x - (4/5)y^3.

How do you find the derivative of e^(-(1/3)x^2-(1/5)y^4)?

To find the derivative of e^(-(1/3)x^2-(1/5)y^4), you need to use the chain rule and the power rule. First, you need to take the derivative of the exponent, which is -(1/3)x^2-(1/5)y^4. Then, you multiply it by the derivative of the inside function, which is e^(-(1/3)x^2-(1/5)y^4). Finally, you multiply it by the derivative of the outside function, which is -2/3x - 4/5y^3.

What is the significance of the negative exponents in the derivative of e^(-(1/3)x^2-(1/5)y^4)?

The negative exponents in the derivative of e^(-(1/3)x^2-(1/5)y^4) indicate that the function is decreasing as x and y increase. This can also be seen in the graph of the function, which has a downward slope.

How does the derivative of e^(-(1/3)x^2-(1/5)y^4) relate to the original function?

The derivative of e^(-(1/3)x^2-(1/5)y^4) is a function that represents the rate of change of the original function. It tells us how fast the original function is changing at a specific point. The derivative can also be used to find the slope of the tangent line at any point on the graph of the original function.

Can the derivative of e^(-(1/3)x^2-(1/5)y^4) be simplified?

Yes, the derivative of e^(-(1/3)x^2-(1/5)y^4) can be simplified to (-2/3)x * e^(-(1/3)x^2-(1/5)y^4) - (4/5)y^3 * e^(-(1/3)x^2-(1/5)y^4), using the power rule and the chain rule. However, it is usually not necessary to simplify the derivative unless explicitly stated in the problem.

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