Solve Derivative: log 5^xsinhx

[ldbaseball16]

Homework Statement

find f '(x) of 5^xsinhx

Homework Equations

i feel like I am missing somthing can someone direct me to the right direction pleasee?

The Attempt at a Solution

product rule (f ')(g)+(f)(g ') f=5^x g=sinhx

5^xsinhx= e^ln(5)^xsinhx

(e^ln(5)^x)(sinhx)+(5^x)(coshx)(ln5)

(e^ln(5)^x)(sinhx)+(5^x)(ln5)(coshx)?

 

[nicksauce]

Well it looks like you took the derivative of none of the terms in the first term, and both of the terms in the second term. Try to fix that.

 

[Unco]

Homework Statement

find f '(x) of 5^xsinhx

Homework Equations

i feel like I am missing somthing can someone direct me to the right direction pleasee?

The Attempt at a Solution

product rule (f ')(g)+(f)(g ') f=5^x g=sinhx

5^xsinhx= e^ln(5)^xsinhx

(e^ln(5)^x)(sinhx)+(5^x)(coshx)(ln5)

(e^ln(5)^x)(sinhx)+(5^x)(ln5)(coshx)?

The factor of ln(5) is in the wrong term.

Write them out separately:
f =
f' =
g =
g' =

Then put them together.

By the way, ensure you use parentheses to remove ambiguity when typing exponents; e.g., 5^x = (e^(ln(5)))^x = \left(e^{\ln{5}}\right)^x etc.

 

[ldbaseball16]

"none of the terms in the first term" what do you mean by this?

 

[Mark44]

Homework Statement

find f '(x) of 5^xsinhx

As Unco pointed out, your expression above is ambiguous. It's not clear what the exponent on 5 is. It could be x, or it could be x*sinh(x). In other words, is this function (5^x)* sinh(x) or is it (5^x)^sinh(x)?

Homework Equations

i feel like I am missing somthing can someone direct me to the right direction pleasee?

The Attempt at a Solution

product rule (f ')(g)+(f)(g ') f=5^x g=sinhx

Now from this work it appears that the function is (5^x) * sinh(x), in other words, as the product of two functions. Inasmuch as you're using the product rule, you must think that the function is (5^x)*sinh(x).

5^xsinhx= e^ln(5)^xsinhx

Now it looks like an exponential function raised to a power, so the function is not a product, which means that the product rule does not apply.

What exactly is the function? We can't help you if you don't know what it is.

(e^ln(5)^x)(sinhx)+(5^x)(coshx)(ln5)

(e^ln(5)^x)(sinhx)+(5^x)(ln5)(coshx)?

 

[HallsofIvy]

5^x is NOT "e^ln(5)^x" it is e^(x ln(5))

Which means that (5^x)'= (e^(x ln(5)))'= ln(5) e^(x ln(5))= ln(5) 5^x.

 

[ldbaseball16]

here is the original problem find f '(x).

f(x)=5^xSinhx does this clarify?

 

[tiny-tim]

oooh, why can't everyone learn how to use the X² tag (just above the Reply box)?

It's e^(ln5)xsinhx.​

 

[ldbaseball16]

yea and that's what i put on top? so since the product rule is (f ')(g)+(f)(g ')
so wouldn't the answer be (e^ln(5)xsinhx+(5^x)(coshx)?

 

[tiny-tim]

yea and that's what i put on top?

nooo! try the X² tag!

so since the product rule is (f ')(g)+(f)(g ')
so wouldn't the answer be (e^ln(5)xsinhx+(5^x)(coshx)?

product rule?

what product?

use the chain rule first! ​

 

[ldbaseball16]

i don't think its the right answer e^sinhx(coshx)

 

[ldbaseball16]

ok so i got 5^xcoshx+5^x(ln5)sinhx?

 

[tiny-tim]

ok so i got 5^xcoshx+5^x(ln5)sinhx?

not even close

show us your full working …

differentiate e^(ln5)xsinhx using the chain rule ​

 

[ldbaseball16]

i used the chain rule and i got e^(ln5xsinhx)(ln5)??

 

[tiny-tim]

what is the derivative of (ln5)xsinhx?

 

[ldbaseball16]

1/5xcoshx

 

[Mark44]

It looks like you are doing this: d/dx(ln 5) = 1/5

If so, that's wrong.

Also, when you differentiate (ln 5)* x*sinh(x), you should be getting two terms.

The answer in your previous post might be interpreted in several ways:
\frac{1}{5x cosh(x)}
\frac{1}{5x}cosh(x)
\frac{1}{5}xcosh(x)

Use parentheses or learn LaTex!

 

[ldbaseball16]

ln5*sinh(x)+ln5*x*cosh(x)?

 

[Mark44]

Yes, that's it. Now you can respond to Tiny Tim's request in post 13.