How do you find the derivative of # log(8x-1)#?

Answer 1

#8/(8x - 1 )#

using the#color(blue)(" chain rule ") #
# d/dx(f(g(x)) = f'(g(x) . g'(x) #

and knowing that #d/dx(logx) = 1/x

# d/dx(log(8x-1)) = 1/(8x-1) d/dx (8x-1) = 8/(8x-1) #
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Answer 2

To find the derivative of log(8x - 1), you can use the chain rule. The derivative of log(u) with respect to x is 1/u times the derivative of u with respect to x. In this case, u = 8x - 1. So, the derivative of log(8x - 1) is 1/(8x - 1) times the derivative of (8x - 1) with respect to x, which is 8. Therefore, the derivative of log(8x - 1) is 8/(8x - 1).

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Answer 3

To find the derivative of log(8x - 1), you can use the chain rule.

The derivative of log(u) with respect to x is 1/u times the derivative of u with respect to x.

So, for log(8x - 1), let u = 8x - 1.

Now, differentiate u with respect to x:

d(u)/dx = d(8x - 1)/dx = 8

Now, apply the chain rule:

d(log(u))/dx = (1/u) * (du/dx)

= (1/(8x - 1)) * 8

= 8/(8x - 1)

Therefore, the derivative of log(8x - 1) with respect to x is 8/(8x - 1).

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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.

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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