Derivatives: If f(x) = 3sin(-2x) / 3 + cos(x), find f'(5)

[mathhelp]

Q1: If f(x) = 3sin(-2x) / 3+cosx
Find f'(5)
Hint: 5 radians

Q2 :f(x) = (4x^2 + x + 7)^cos(x)
Find f'(x)
Find f'(1)
Hint: logarithmic differentiation

I would appreciate your solutions to the above problems.

 

[soroban]

Re: Derivatives

Hello, mathhelp!

These are challenging problems in differentiating.
But I don't see the point of plugging in 5 or 1 into the results.

Also, exactly where is your difficulty?

\(\displaystyle 1)\L\;f(x)\:=\:\frac{3\sin(-2x)}{3\,+\,\cos(x)}\). . Find $\displaystyle \,f'(5)$

Quotient Rule:

. . $\displaystyle f'(x)\;=\;\frac{[3\,+\,\cos(x)]\cdot[3\cos(-2x)\cdot(-2)] \,-\,[3\sin(-2x)]\cdot[-\sin(x)]}{[3\,+\,\cos(x)]^2}$

. . \(\displaystyle \fbox{ f'(x) \;= \;\frac{-6\cos(-2x)\cdot[3\,+\,\cos(x)]\,+\,3\sin(x)\cdot\sin(-2x)}{[3\,+\,\cos(x)]^2}}\)

Now plug in $\displaystyle x\,=\,5$

\(\displaystyle 2)\;f(x) \:= \4x^2\,+\,x\,+\,7)^{\cos(x)}\). . Find $\displaystyle f'(x)$ and $\displaystyle f'(1)$

We have: \(\displaystyle \:y\;=\;\left(4x^2\,+\,x\,+\,7)^{\cos x}\)

Take logs: $\displaystyle \:\ln(y)\;=\;\ln\left(4x^2\,+\,x\,+\,7\right)^{\cos x} \;= \;\cos x\cdot\ln\left(4x^2\,+\,x\,+\,7\right)$

Differentiate implicitly:
. . \(\displaystyle \frac{1}{y}\cdot y' \;= \;\cos x\L\cdot\frac{1}{4x^2\,\)\(\displaystyle +\,x\,+\,7}\cdot(8x\,+\,1)\,+\,(-\sin x)\cdot\ln\left(4x^2\,+\,x\,+\,7\right)\)

. . $\displaystyle \frac{y'}{y} \;= \;\frac{(8x\,+\,1)\cos x}{4x^2\,+\,x\,+\,7}\,-\,\sin x\cdot\ln\left(4x^2\,+\,x\,+\,7\right)$

Multiply by $\displaystyle y:\;\;y' \;=\;y\cdot\left[\frac{(8x\,+\,1)\cos x}{4x^2\,+\,x\,+\,7}\,-\,\sin x\cdot\ln\left(4x^2\,+\,x\,+\,7\right)\right]$

Replace \(\displaystyle y:\;\;\fbox{f'(x) \;=\;(4x^2\,+\,x\,+\,7)^{\cos x}\cdot\left[\frac{(8x\,+\,1)\cos x}{4x^2\,+\,x\,+\,7}\,-\,\sin x\cdot\ln\left(4x^2\,+\,x\,+\,7\right)\right]}\)

Now let $\displaystyle x\,=\,1$

 

[mathhelp]

difficulty

What would be the result of cos(-10) if I was to plug 5 in the derivative?

 

[soroban]

Re: Derivatives

Hello, mathhelp!

We had: \(\displaystyle \L\:f'(x) \;= \;\frac{-6\cos(-2x)\cdot[3\,+\,\cos(x)]\,+\,3\sin(x)\cdot\sin(-2x)}{[3\,+\,\cos(x)]^2}\)

We're supposed to let $\displaystyle x\,=\,5.$


I don't understand your difficulty . . .
Are you having trouble entering it into your calculator?

You ask "What is cos(-10) ?"

First, be sure your calculator is in radian mode.

Then enter: \(\displaystyle \,\fbox{\cos}\;\fbox{(-)}\;\fbox{1}\;\fbox{0}\;\fbox{=}\)

You should get: \(\displaystyle \L\:\fbox{-0.839071529}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

That $\displaystyle \fbox{(-)}$ is not the subtraction key.

You should have a key which changes the sign of a number.

On some calculators, it looks like this: $\displaystyle \,\fbox{+/-}$