1. ## differentiate!

1) $f(x)=(7x+\sqrt{x^2+6})^4$

2) $f(x)=tan^3 (4x)$

3) $g(x)=(\frac{x-2}{2x+2})^9$

4) $p(x)= [3x(5x-1)^5]^4$

I tried to work out all 4 of these but I keep getting the wrong answer.

2. Originally Posted by yoman360
1) $f(x)=(7x+\sqrt{x^2+6})^4$

2) $f(x)=tan^3 (4x)$

3) $g(x)=(\frac{x-2}{2x+2})^9$

4) $p(x)= [3x(5x-1)^5]^4$

I tried to work out all 4 of these but I keep getting the wrong answer.
all involve the chain rule ...

1) $f'(x) = 4(7x + \sqrt{x^2+6})^3 \cdot \left(7 + \frac{x}{\sqrt{x^2+6}}\right)$

2) $f'(x) = 3\tan^2(4x) \cdot \sec^2(4x) \cdot 4$

3) $g'(x) = 9\left(\frac{x-2}{2x+2}\right)^8 \cdot \frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}$

4) $p'(x)= 4[3x(5x-1)^5]^3 \cdot [3x \cdot 5(5x-1)^4 \cdot 5 + (5x-1)^5 \cdot 3]$

you can clean up the algebra.

3. Originally Posted by skeeter
all involve the chain rule ...

1) $f'(x) = 4(7x + \sqrt{x^2+6})^3 \cdot \left(7 + \frac{x}{\sqrt{x^2+6}}\right)$

2) $f'(x) = 3\tan^2(4x) \cdot \sec^2(4x) \cdot 4$

3) $g'(x) = 9\left(\frac{x-2}{2x+2}\right)^8 \cdot \frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}$

4) $p'(x)= 4[3x(5x-1)^5]^3 \cdot [3x \cdot 5(5x-1)^4 \cdot 5 + (5x-1)^5 \cdot 3]$

you can clean up the algebra.
Thanks I kinda don't understand the chain rule but as I do more problems it might make more sense.