Thread: 2 Questions Derivative

Two questions:

#1:



(fog)'(3)=?
(gof)'(3)=?

#2:

Let F(x)= f(x^{6}) and G(x)=(f(x))^{6} . You also know that a^{5}=10, f(a)=2,f'(a)=12, f'(a^{6})=10.

Find F'(a)=

Originally Posted by qbkr21

Two questions:

#1:



(fog)'(3)=?
(gof)'(3)=?

#2:

Let F(x)= f(x^{6}) and G(x)=(f(x))^{6} . You also know that a^{5}=10, f(a)=2,f'(a)=12, f'(a^{6})=10.

Find F'(a)=

#1 I'd like to help, but I'm not sure what the notation (fog)'(3) means. Is the ' a prime, representing a derivative?

#2 Find F'(a):
Since F(x) = f(x^6), taking the derivative, we need to use chain rule:
F'(x) = f'(x^6)*6x^5
F'(a) = f'(a^6)*6a^5

Now, we know f'(a^{6})=10, a^{5}=10, so we get
F'(a) = {10}*6*{10} = 600

F composed of g like a composition...



Also I have another question...


If I wanted to differentiate

-2y

for implicit differentiation would the derivative be

-2 or

-2yy^prime




Thanks so much

Originally Posted by qbkr21

F composed of g like a composition...



Also I have another question...


If I wanted to differentiate

-2y

for implicit differentiation would the derivative be

-2 or

-2yy^prime




Thanks so much

-2 y'

that is if you're differentiating with respect to a variable other than y

Originally Posted by qbkr21

F composed of g like a composition...



Also I have another question...


If I wanted to differentiate

-2y

for implicit differentiation would the derivative be

-2 or

-2yy^prime




Thanks so much

d/dx(-2y) = -2*dy/dx = -2*y' <-- the ' is "prime"

Thanks!

I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

y^3 so 3y^2y'

another

2y^2 so 4yy'

again

-2y so -2y'

is this correct... if so I am set!!!

Originally Posted by qbkr21

Two questions:

#1:



(fog)'(3)=?
(gof)'(3)=?

Find F'(a)=

(fog)(3) = f(g(3))
First we need the value of g at x=3: g(3) = 4. To get this, just look on the x-axis where x=3 and move up until you hit the graph of g, the y-value at that point is g(3)=4.
Second, we need f(g(3)) = f(4) = 2. Use the same method for x=4, going up to the graph of f. The y-value at that point is f(4)=2.

Now, do the same for (gof)(3) = g(f(3)).

Originally Posted by qbkr21

I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

y^3 so 3y^2y'

another

2y^2 so 4yy'

again

-2y so -2y'

is this correct... if so I am set!!!

correct

Originally Posted by ecMathGeek

(fog)(3) = f(g(3))
First we need the value of g at x=3: g(3) = 4. To get this, just look on the x-axis where x=3 and move up until you hit the graph of g, the y-value at that point is g(3)=4.
Second, we need f(g(3)) = f(4) = 2. Use the same method for x=4, going up to the graph of f. The y-value at that point is f(4)=2.

Now, do the same for (gof)(3) = g(f(3)).

so we're forgetting about the primes? i hope we can do that

Thanks you guys. YOU ARE THE BEST!!!

Still confused a bit

(f composed of g)' is not equal to 2 according to the computer...

Originally Posted by qbkr21

I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

y^3 so 3y^2y'

another

2y^2 so 4yy'

again

-2y so -2y'

is this correct... if so I am set!!!

The technical rule is that any time you 'take the derivative of y' you get a y'. For example, let's say I'm taking the derivative of 2xy implicitly. This is product rule, so: D(2xy) = 2x*y' + 2y. When I took the derivative of y, I got y', when I took the derivative of x, 1.

Besides the technical rule, what you have done is correct.

Originally Posted by Jhevon

so we're forgetting about the primes? i hope we can do that

That's why I asked about the ' . I'm not sure what to do with the prime there.

Does anyone know how to solve question #1 with the graph? After studying the picture for quite some time I have come to the conclusion that the lines on the graph are f(x) and g(x), they are not f'(x) and g'(x). Any assistance would wonderful. This is my last problem for Webwork and I am clueless as to what I should do.

Thanks!