Verification of answer

[sinclairharry]

Question is to derive 5x ^ 6x

I believe the answer to be 30x ^ (6x) ( ln (5x) + 1)

Is it correct?

 

[Deleted member 4993]

Question is to derive 5x ^ 6x

I believe the answer to be 30x ^ (6x) ( ln (5x) + 1)

Is it correct?

Is your question:

Calculate the derivative, with respect to x, of
y = 5x^6x

 

[sinclairharry]

Is your question:

Calculate the derivative, with respect to x, of
y = 5x^6x

yes !

 

[ksdhart]

I worked the problem and my answer is very similar to yours, except the natural log is of x instead of 5x. I'll go through my steps.

Starting with \(\displaystyle y = 5x^{6x}\)

Divide by 5: \(\displaystyle \frac{y}{5}=x^{6x}\)

Take the natural log: \(\displaystyle ln\left(\frac{y}{5}\right)=ln\left(x^{6x}\right)\)

Apply the Power Rule for Logarithms: \(\displaystyle ln\left(\frac{y}{5}\right)=6x\cdot ln\left(x\right)\)

Take the derivative: \(\displaystyle \frac{d}{dx}\left(ln\left(\frac{y}{5}\right)\right)=\frac{d}{dx}\left(6x\cdot ln\left(x\right)\right)\)

Apply Implicit Differentiation and Product Rule for Derivatives: \(\displaystyle \frac{y'\left(x\right)}{y\left(x\right)}=6\cdot ln\left(x\right)\:+\:6x\cdot \frac{1}{x}\)

Simplify: \(\displaystyle \frac{y'\left(x\right)}{y\left(x\right)}=6\left(ln\left(x\right)+1\right)\)

Multiply by y(x) to get the final answer: \(\displaystyle y'(x)=30x^{6x}\left(ln\left(x\right)+1\right)\)

I hope that seeing the process I used was helpful to you.

 

[sinclairharry]

nope, i was correct !

 

[Ishuda]

nope, i was correct !

No. You are not correct for either
y=(5x)^6x
or
y=5(x)^6x
For the first you are off by a factor of 5 and for the second the ln(5x) should be ln(x) as ksdhart demonstrated. I would do it a slightly different way if I didn't know the formula; write the first as
y = e ^ln(y) = e^{6x ln(5x)}
and
y' = e^{6x ln(5x)}(6x ln(5x))'
= (5x)^6x (6x ln(5x))'
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{(5x)'}{5x}\)]
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{5}{5x}\)]
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{1}{x}\)]
= (5x)^6x [6 ln(5x) + 6]
= 6 (5x)^6x [ln(5x) + 1]

 

[sinclairharry]

No. You are not correct for either
y=(5x)^6x
or
y=5(x)^6x
For the first you are off by a factor of 5 and for the second the ln(5x) should be ln(x) as ksdhart demonstrated. I would do it a slightly different way if I didn't know the formula; write the first as
y = e ^ln(y) = e^{6x ln(5x)}
and
y' = e^{6x ln(5x)}(6x ln(5x))'
= (5x)^6x (6x ln(5x))'
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{(5x)'}{5x}\)]
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{5}{5x}\)]
= (5x)^6x [6 ln(5x) + 6x \(\displaystyle \frac{1}{x}\)]
= (5x)^6x [6 ln(5x) + 6]
= 6 (5x)^6x [ln(5x) + 1]

Wouldn't 6 (5x)^6x [ln(5x) + 1] be equal to 30x^6x [ln(5x) + 1] by simply multiplying the 6 by 5. Meaning my answer is correct ?

 

[daon2]

Wouldn't 6 (5x)^6x [ln(5x) + 1] be equal to 30x^6x [ln(5x) + 1] by simply multiplying the 6 by 5. Meaning my answer is correct ?

no, for the same reason 2*3^2 = 18 is not the same as 6^2=36.