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1)F(x) ex2 +sinx

2)g(x) = sin(x2) under root

3) 3x2+5
h(x) log 4 ( log to the bese 4) 3x square +5

2006-12-25 23:49:23 · 3 answers · asked by astraelk 1 in Science & Mathematics Mathematics

3 answers

I'm not sure what "all procedures" means, but...

1) if f(x) = e^(x^2) + sinx
then f'(x) =2x e^(x^2) + cos(x) (using the chain rule on the first part)

2) if g(x) = sqrt(sin(x^2)) then we need to use the chain rule twice:
g'(x) = .5 (sin(x^2))^-.5 * cos(x^2) * 2x
which I'll let you simplify by combining terms like .5 *2...

3) h(x) = logbase4(3x^2+5)
= logbasee(3x^2+5)/logbasee(4)
h'(x) = 1/logbasee(4) * 1/(logbase3(3x^2+5) * 6x

2006-12-26 00:42:20 · answer #1 · answered by firefly 6 · 0 0

1) F(x) = e^(x^2) + sin(x)

Note that the derivative of e^x is itself, e^x. Also, the derivative of sin(x) is cos(x). The derivative of x^2 is 2x (this is where we use the chain rule. Therefore,

F'(x) = e^(x^2) {2x} + cos(x)

{I denoted the use of the chain rule using the { } brackets.}

2) g(x) = sqrt (sin(x^2))

Note that taking the square root is the same as taking to the power of (1/2). So we can rewrite g(x) as

g(x) = [sin(x^2)]^(1/2).

Here, we're making use of taking the derivative of three functions, in these forms: z^(1/2), sin(z), and z^2. We apply the chain rule twice.

g'(x) = (1/2) [sin(x^2)]^(-1/2) {cos(x^2)} {2x}

3) This question is incomprehensible so I can't do it.

2006-12-26 09:44:58 · answer #2 · answered by Puggy 7 · 0 0

please clarify whether it is e^x^2+sinx?
2.whether it is [sin(x^2)]^1/2?
3.whether it is log (3x^2+5) base 4
what is h(x)log4(to the base 4)3x squared +5?
please repost the sums exactly as they are in the text book

2006-12-26 08:42:48 · answer #3 · answered by raj 7 · 0 0

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