Hey you all . Help me if you're not too busy (: Calc problem

[PrincessOfSteel92]

1.) Create a formula for d/dx (f (x) . g(x) . h(x)) . Then from that use that formula to find d/dx (3x[sup:1nq7tzqp]2[/sup:1nq7tzqp]e[sup:1nq7tzqp]x[/sup:1nq7tzqp]sinx)

The problem where I get stuck here is how I go about finding a formula out of the first equation and more-so using it to find the second equations derivative.






2.) create a formula for d/dx (f o g o h (x))

Then use it to find d/dx (Sin e [sup:1nq7tzqp]3x^2][/sup:1nq7tzqp]



My main prblems with these two is going about finding the 'formula'. No idea what the book means by that, and for the second one I totally blanked out about what o means.

Thanks so much, please help.

 

[galactus]

Re: Hey you all . Help me if you're not too busy Calc pro

1.)

Create a formula for d/dx (f (x) . g(x) . h(x)) . Then from that use that formula to find d/dx (3x[sup:2b072fvo]2[/sup:2b072fvo]e[sup:2b072fvo]x[/sup:2b072fvo]sinx)

They're asking for the product rule for three terms.

\(\displaystyle (f\cdot g\cdot h)'(x)=[(f\cdot g)\cdot h]'(x)\)

\(\displaystyle =(f\cdot g)(x)h'(x)+h(x)(f\cdot g)'(x)\)

\(\displaystyle =(f\cdot g)(x)h'(x)+h(x)[f(x)g'(x)+f'(x)g(x)]\)

\(\displaystyle =f(x)g(x)h'(x)+f(x)g'(x)h(x)+f'(x)g(x)h(x)\)

See the pattern?. You can even come up with a formula for n products. Say, for 5 terms. Whatever.

Now, use it to find the derivative of \(\displaystyle 3x^{2}\cdot e^{x}\cdot sin(x)\)

The second problem is asking for \(\displaystyle \frac{d}{dx}[f(g(h(x)))]=\frac{d}{dx}[f(g(u))], \;\ u=h(x)\).

Make a substitution. It is the chain rule. Continue?.

 

[Queenisabella87]

Re: Hey you all . Help me if you're not too busy Calc pro

Oh thanks for your explaination!

I have a question though, how does the pattern come into solving for the equation ?

And where do i use the chain rule? for the 2nd problem