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How do you differentiate #e^(-10x)#?

Answer 1

Scarlett Allison

#-10e^(-10x)#

Given: #e^(-10x)#.

Use the chain rule, which states that,

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

Let #u=-10x,:.(du)/dx=-10#.

Then, #y=e^u,:.dy/(du)=e^u#.

Combining, we get:

#dy/dx=e^u*-10#

#=-10e^u#

Substitute back #u=-10x# to get the final answer:

#=-10e^(-10x)#

Note:

A common fact in derivatives is that #d/dx(e^(f(x)))=f'(x)e^(f(x))#.

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Answer 2

Ezekiel Alexander

#d/dx(e^(-10x))=-10*e^(-10x)#

By the formula #(d(e^u))/dx=e^u*(du)/dx#

#d/dx(e^(-10x))=e^(-10x)*(d(-10x))/dx#

#d/dx(e^(-10x))=e^(-10x)*(-10)#

#d/dx(e^(-10x))=-10*e^(-10x)#

I hope the explanation is useful....God bless...

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Answer 3

Adam Adkins

To differentiate e^(-10x), you can use the chain rule. The derivative of e^u with respect to x is e^u times the derivative of u with respect to x. In this case, u = -10x. So, the derivative is e^(-10x) times the derivative of -10x with respect to x, which is -10. Therefore, the derivative of e^(-10x) is -10e^(-10x).

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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.