# Math Help - chain rule problems

1. ## chain rule problems

I have been having trouble with the ADVANCED portion of the chain rule and I would like some help on these.
Find the derivative of these equations
1. (sinx^2)(cosx^2) The answer comes out to be 2x cos2x^2
2. (x^3)(2-3x)^2. I know the first is 3x^2 and for the deriv. of the other one, it would be 2(2-3x)(2-3x)'-> (3x^2)(2(2-3x)-3). But the answer is 3x^2(2-3x)(2-5x).
3.(sqrt x^2+3/x^2-5) The answer comes out to (1/2)(x^2+3/x^2-5)^-1/2(-16x/(x^2-5)^2)

2. Originally Posted by driver327
I have been having trouble with the ADVANCED portion of the chain rule and I would like some help on these.
Find the derivative of these equations
1. (sinx^2)(cosx^2) The answer comes out to be 2x cos2x^2)
$\sin x^2 \cos x^2 = \frac{1}{2} \sin (2x^2)$
Derivative,
$\frac{1}{2} \cdot (2x) \cos (2x^2) = x\cos (2x^2)$

3. Originally Posted by driver327
2. (x^3)(2-3x)^2. I know the first is 3x^2 and for the deriv. of the other one, it would be 2(2-3x)(2-3x)'-> (3x^2)(2(2-3x)-3). But the answer is 3x^2(2-3x)(2-5x).
$(x^3(2-3x)^2)^{\prime} = (x^3)^{\prime}(2 - 3x)^2 + x^3 ((2 - 3x)^2)^{\prime}$

Now,
$(x^3)^{\prime} = 3x^2$
and
$((2 - 3x)^2)^{\prime} = 2(2 - 3x) \cdot -3 = -6(2 - 3x)$

So

$(x^3(2-3x)^2)^{\prime} = (x^3)^{\prime}(2 - 3x)^2 + x^3 ((2 - 3x)^2)^{\prime}$ $= 3x^2 \cdot (2 - 3x)^2 + x^3 \cdot -6(2 - 3x)$

You can simplify it from here.

-Dan

4. Originally Posted by driver327
3.(sqrt x^2+3/x^2-5) The answer comes out to (1/2)(x^2+3/x^2-5)^-1/2(-16x/(x^2-5)^2)
Please use parenthesis. I presume the problem is to find the derivative of
$\sqrt{\frac{x^2 + 3}{x^2 - 5}}$

So
$\left ( \sqrt{\frac{x^2 + 3}{x^2 - 5}} \right ) ^{\prime} = \left ( \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{1/2} \right ) ^{\prime}$

$= \frac{1}{2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{-1/2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{\prime}$

Now,
$\left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{\prime} = \frac{2x \cdot (x^2 - 5) - (x^2 + 3) \cdot 2x}{(x^2 - 5)^2}$

$= \frac{2x^3 - 10x - 2x^3 - 6x}{(x^2 - 5)^2} = \frac{-16x}{(x^2 - 5)^2}$

So finally:
$\left ( \sqrt{\frac{x^2 + 3}{x^2 - 5}} \right ) ^{\prime} = \frac{1}{2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{-1/2} \cdot \frac{-16x}{(x^2 - 5)^2}$

Again, I'll let you simplify it from here.

-Dan