Derivatives: If f(x) = 3sin(-2x) / 3 + cos(x), find f'(5)
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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?
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 \(\displaystyle x\,=\,5\)
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 \(\displaystyle x\,=\,1\)
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?
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 \(\displaystyle x\,=\,1\)
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: \(\displaystyle \,\fbox{+/-}\)
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: \(\displaystyle \,\fbox{+/-}\)