for (a) use. when y=uv
Just use the product rule, as arze says.
No. The formula requires the that base be the variable x and the exponent be a constant. Here it is the other way around. The derivative of is .b) f(x)=3^x+x^3
f'(x)=x(3)^x-1 + 3X^2
PLEASE don't use "x" to mean the variable and to mean times!c) y=e^-(2x+5)
=e^-(2x+5) x ln e x -2
The derivative is, indeed, e^-(2x+5)(-2)(ln e), but ln e= 1. This is just -2e^-(2x+ 5).
Again, don't mix "x" as variable and "times"! Also ln e= 1. This is just (6x-5)e^(3x^2- 5x+ 7).d) y=e^(3x^2 -5x+7)
=e^(3x^2 -5x+7) x ln e x 6x-5