What are the formulas for the derivatives of e^x and ln(x)? How do the formulas

Question

What are the formulas for the derivatives of e^x and ln(x)?

How do the formulas change when the base of a logarithm is other than e?

How can we use these formulas to determine the derivatives of more complex functions involving e^x and ln(x)?

What role do the rules for expanding logarithmic functions play in making it easier to find derivatives of complex logarithmic functions?

What is L'Hospitals Rule? What conditions must be met in order to make use of it? How do we use it to find a limit?

What real-world applications can you find involving exponential and logarithmic functions? Explain how you would use derivatives to explore those applications.

Explanation / Answer

1) What are the formulas for the derivatives of e^x and ln(x)?

Ans :

Derivative of e^x is : e^x itself

Derivative of ln(x) is : 1 / x

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2) How do the formulas change when the base of a logarithm is other than e?

Ans :

When it is osme other base, we need to first convertit into ln using change of base rule.
For example log (base 2) x = ln(x) / ln(2), where 1/ln(2) is a constant.
So, if we wanna derive log (base 2) x, then it would be : (1/ln2) * (1/x)

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3) How can we use these formulas to determine the derivatives of more complex functions involving e^x and ln(x)?

Ans :

For example, say we wanna find the derivative of e^(3x)
d/dx(e^(3x)) --> Since this is of the form e^u, the result of deriving would be e^u itself
e^(3x)
But by Chain Rule, we have to drill down, as in multiply the derivative of the exponent to e^(3x)
Derivative of 3x is : 3
So, final ans : e^(3x) * 3
3e^(3x)
Chain Rule is ALWAYS employed when we find the derivative of complex functions, even when finding derivatives of seemingly simple functions like sin(x+2) or cos(2x - 1) is a usage of chain rule of derivatives

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4) What role do the rules for expanding logarithmic functions play in making it easier to find derivatives of complex logarithmic functions?

Ans :

Say we have to find the derivative of (x+1)^2*(x-2)^2 / (x - 3)^2.
Now, instead of expanding numerator and deriving using quotient rule or directly deriving using quotient rule, we can use the log method. When we "ln" both sides, we get :

ln [ (x+1)^2*(x-2)^2 / (x - 3)^2 ] on the right
Since this is of the form ln(A/B), which is ln(A) - ln(B), this becomes :

ln[(x+1)^2 * (x-2)^2] - ln[(x-3)^2]

Now, the firts ln term is of the form ln(a*b), which is lna + lnb

So, it becomes :

ln((x+1)^2) + ln((x-2)^2) - 2*ln(x-3)

2ln(x+1) + 2ln(x-2) - 2ln(x-3)

Now, when we derive, we get :

2/(x+1) + 2/(x-2) - 2/(x-3), a much simpler looking result

This is the reason.

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5) What is L'Hospitals Rule? What conditions must be met in order to make use of it? How do we use it to find a limit?

Ans :

L Hospital's rule is a technique of finding limits by deriving the numerator and denominator separately as a means of simplifying the process of finding the limit. The conditions to be met is that when we plug in the limit value into the given function, it must be of the 0/0 or infinity/infinity form.

Say we have to find the limit :

lim x --> 2 (x^2 - 4) / (x - 2)

When you plug in limit, x = 2, we get (2^2 - 4) / (2 - 2) ---> (4 - 4) / 0 ---> 0/0 form

so, deriving using L Hospital;'s rule, the limit now becomes :

lim x ---> 2 2x / 1

lim x --> 2 (2x)

Now, applying the limit, we get 2(2) ---> 4

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