Derivatives of exponent x with a product

In summary, the person is trying to find the derivative of f '(1), but keeps getting incorrect answers. They try a different approach and get the same incorrect answer. They try plugging in f '(1) into the equation and get an answer that is very close to the actual answer.
  • #1
cal.queen92
43
0

Homework Statement



f(x) = 10(sin(x))^x ----> find f '(1)


The Attempt at a Solution



I have tried several different approaches, but still get stuck with a wrong answer every time

f(x) = 10(sin(x))^x let f(x) = y so y=10(sin(x))^x then ln y = ln10(sin(x))^x

using log. laws: lny = xln10(sin(x))

then differentiating implicitly using product and chain rule:

1/y*dy/dx = ln10(sin(x)) + 1/10(sinx) * 10(cosx)*x so

1/y*dy/dx = (ln10(sinx)) + (x(cosx)/(sinx))

then multiplying both sides by y to eliminates denominator:

dy/dx = ((10(sinx))^x)*((ln10(sinx))+(x(cosx)/(sinx))

^ this would be the unsimplified derivative ^

Now for f '(1) I would just plug in 1 where ever there is an x right?

I am doing something wrong, can anyone see my mistake? Thank you!
 
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  • #2
[tex] \log (10 (\sin x)^x) \neq x \log (10 \sin x) [/tex]
[tex] \log (10^x (\sin x)^x) = x \log (10 \sin x) [/tex]
 
  • #3
Hmmm I see, so then what if i try a different approach:

lny = ln10 + xln(sinx)

I continued this way but got the wrong answer again. Is this a logical way of approaching it?
 
  • #4
That should work. What did you get for the derivative after making that change?
 
  • #5
If I continue, I get:

(1/y)(dy/dx) = ((1/10)+((1/(sinx))(cosx)))

then

(1/y)(dy/dx) = ((1/10)+(cotx))

Then multiplying both sides by y:

dy/dx = (10(sin(x))^1)((1/10)+(cotx))

If I fill in the f(1) then it looks like:

dy/dx = (10(sin(1))^1)((1/10)+(cot(1))

But, another thing that puzzles me is that on a unit circle, Sin(1) = 0 right? So wouldn't the entire answer turn to 0 sine everything is multiplied by Sin(1)?
 
  • #6
[itex]\log 10[/itex] is a constant, not a function of x. You still need to use product and chain rules when differentiating [itex] x \log (\sin x) [/itex].

And, no, [itex] \sin 1 [/itex] isn't 0. [itex]\sin 0, \sin \pi, \sin 2\pi[/itex], etc, are zero.
 
  • #7
Okay, so if ln10 is a constant than its derivative goes to zero. Now I am getting an answer that looks like:

lny = ln10 + xln(sinx)

(1/y)(dy/dx) = 1*ln(sinx) + (1/(sinx))(cosx)(x)

(1/y)(dy/dx) = ln(sinx) + (xcosx/(sinx))

then multiplying both sides by y:

(dy/dx) = (ln(sinx) + (xcosx/(sinx)))(10(sinx)^x)

Then multiplying through:

(10ln(sinx)^(x+1)) + ((10x(cosx)(sinx)^x)/(sinx))

Which is actually very close to the answer:

(10(sinx)^(x+1))(ln(sinx))+(10x(cosx)(sinx)^x))/(sinx)

But these answers aren't exactly the same, I must be missing something.
 
  • #8
cal.queen92 said:
then multiplying both sides by y:

(dy/dx) = (ln(sinx) + (xcosx/(sinx)))(10(sinx)^x)

Then multiplying through:

(10ln(sinx)^(x+1)) + ((10x(cosx)(sinx)^x)/(sinx))

ln(sin(x)) and 10(sin x)^x can't be multiplied that way. ln (sin x)*10 (sin x)^x is simplified already.
 
  • #9
Ohhhh okay, so now the answer looks like:

((ln(sinx))(10(sinx)^(x+1)))+((10x(cosx)(sinx)^x)/(sinx))

But in the actual answer, the whole thing is over sinx, where this answer isn't, it's so close!
 
  • #10
Well you left the x+1 in the exponent. So you'd need a sin x in the denominator to make that right.
 
  • #11
Okay, so if I did not leave the x+1 in the exponent then the sinx in the denominator wouldn't be necessary? Thanks for all your help!
 

Related to Derivatives of exponent x with a product

1. What is the formula for finding the derivative of a product of two functions with an exponent of x?

The formula for finding the derivative of a product of two functions with an exponent of x is (f(x)g(x))' = f'(x)g(x) + f(x)g'(x).

2. How do you simplify a derivative of a product with an exponent of x?

To simplify a derivative of a product with an exponent of x, you can use the product rule and the power rule. First, use the product rule to find the derivative of the product. Then, use the power rule to simplify the terms with an exponent of x.

3. Can you find the derivative of a product with an exponent of x using the quotient rule?

No, the quotient rule is used to find the derivative of a quotient, not a product. To find the derivative of a product with an exponent of x, you must use the product rule.

4. What is the difference between finding the derivative of a product with an exponent of x and finding the derivative of a product without an exponent?

The difference is that when finding the derivative of a product with an exponent of x, you must use the product rule, and when finding the derivative of a product without an exponent, you can simply distribute the derivative to each term in the product.

5. How do you find the derivative of a product with multiple exponents of x?

To find the derivative of a product with multiple exponents of x, you can use the product rule and the power rule. First, use the product rule to find the derivative of the product. Then, use the power rule to simplify each term with an exponent of x.

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