# Any help on these 2 questions would be greatly appreciated, I kinda know the differentiantion question but just wanna get the answer! thank you

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For f) we will graph a=1 (black), a=2 (red), and a=3 (orange) to show what happens to the graph as a increases:

As you can see, as a increases the concavity of the function becomes tighter.  Furthermore, the sign of the second derivative of a function can tell you it's concavity (positive for concave up):

`f(x)=a^x`

`f'(x)=a^xln(a)`

`f''(x)=a^xln(a)^2`

`a^xgt0` because a>0 and `ln(a)^2gt0`

Therefore, because f''(x)>0 the graph is concave up.

f) `f(x)=sin(x^2+2x)sqrt(2x-7)` find f'(x)

Use the product rule:

`u(x)=sin(x^2+2x) -gt u'(x)=(2x+2)cos(x^2+2x)`

`v(x)=sqrt(2x-7)-gtv'(x)=1/2(2)(1/sqrt(2x-7))=1/(sqrt(2x-7))`

`f'(x)=u(x)v'(x)+v(x)u'(x)`

`=sin(x^2+2x)(1/(sqrt(2x-7)))+(sqrt(2x-7))(2x+2)cos(x^2+2x)`

`=(sin(x^2+2x)+(2x-7)(2x+2)cos(x^2+2x))/sqrt(2x-7)`

`=(sqrt(2x-7)sin(x^2+2x)+sqrt(2x-7)(4x^2-10x-14)cos(x^2+2x))/(2x-7)`