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Could you check my calculation of derivative for this Problem..

Y= 7x(Cosx)^(x/2)

Q. FIND DY/DX

1- Ln both sides

Working on the RHS first

Ln(7x(Cosx)^(x/2))

2. Then "Expand the bracket" using logaritmic rules=

Ln(7x) +Ln(Cosx)^(x/2)

3. Now, Bring down the power in the second term to the front

Ln(7x) + (x/2)Ln Cosx

4. Differenriate as usual with respect to x

First term

Ln(7x) = 1/(7x)

For the Second term

(X/2) LnCos X

Use the product rule and the chain rule because there are 2 functions being multiplied and an inner and outer function.

Using the product rule to differentiate (x/2)Ln Cosx

U= x/2 V= lnCosx

U'v +V'u=

(X/2) (LnCosx')+ LnCosx(X/2')

Differentiating LnCos x requires the Chain rule

= (1/Cosx)* -Sinx
Since sinx/cosx = Tan x, -(sinx/cosx) = -tanx

second term (of the product rule) - derivative of x/2= 1/2


So together we have:

(X/2)*-tanx+LnCosx(1/2)

Simplifying:

(-tanx+lncosx)/2

4. Working on the LHS now

Lny=( 1/y)dy/dx

Therefore the whole thing is:

(1/Y)dy/dx= -Tanx+Lncosx/2

Multiplying both sides by y we get:

Dy/dx= Y *( -Tanx+Lncosx/2)

Substituting the value of Y from the equation

Recall: Y= 7x(cosx)^(x/2)

Therefore dy/dx=

7x(cosx)^(x/2)* ((-Tanx+Lncosx)/2)



Please can anyone check this for me, I would be ever so grateful!! 😊😊😊
Original post by FloralEssence
...


No that's not right. Also PLEASE use LaTex or clearer representation, there's a lot of ambiguity there.

ddx(ln⁡∣7x∣)=1x≠17x\frac{d}{dx}(\ln \lvert 7x \lvert )=\frac{1}{x}\not= \frac{1}{7x}

and when you say "the whole thing is..." you are missing the derivative I corrected above.

ddx(x2ln⁡∣cos⁡x∣)\frac{d}{dx}(\frac{x}{2} \ln\lvert \cos x \lvert ) is correct though.

ALSO please do not say ln⁡(7x)=17x\ln(7x) = \frac{1}{7x} as it doesn't make sense when you're actually differentiating it...

(the more i look through this the more errors i seem to find)
(edited 7 years ago)

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