7. (2 points) find the derivative of each function. pick 2.\na. $f(x)=(3x^{2}-4x + 5)cdot2^{x}$\nb…

1. For function (f(x)=(3x^{2}-4x + 5)\cdot2^{x}):
   • Step 1: Recall the product - rule
     • The product - rule states that if (y = u\cdot v), then (y^\prime=u^\prime v+uv^\prime). Let (u = 3x^{2}-4x + 5) and (v = 2^{x}).
     • First, find (u^\prime):
       • Using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (u^\prime=(3x^{2}-4x + 5)^\prime=6x-4).
     • Then, find (v^\prime):
       • The derivative of (a^{x}) with respect to (x) is (a^{x}\ln a), so (v^\prime=(2^{x})^\prime=2^{x}\ln 2).
   • Step 2: Apply the product - rule
     • (f^\prime(x)=u^\prime v+uv^\prime=(6x - 4)\cdot2^{x}+(3x^{2}-4x + 5)\cdot2^{x}\ln 2).
     • Factor out (2^{x}): (f^\prime(x)=2^{x}[(6x - 4)+(3x^{2}-4x + 5)\ln 2]=2^{x}(6x - 4 + 3x^{2}\ln 2-4x\ln 2 + 5\ln 2)).
2. For function (g(x)=\sqrt{x}\cdot(\ln 3)^{x}):
   • Step 1: Recall the product - rule
     • Let (u=\sqrt{x}=x^{\frac{1}{2}}) and (v = (\ln 3)^{x}).
     • First, find (u^\prime):
       • Using the power - rule ((x^n)^\prime=nx^{n - 1}), we get (u^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}).
     • Then, find (v^\prime):
       • Since the derivative of (a^{x}) with respect to (x) is (a^{x}\ln a), (v^\prime=(\ln 3)^{x}\ln(\ln 3)).
   • Step 2: Apply the product - rule
     • (g^\prime(x)=u^\prime v+uv^\prime=\frac{1}{2\sqrt{x}}\cdot(\ln 3)^{x}+x^{\frac{1}{2}}\cdot(\ln 3)^{x}\ln(\ln 3)).
     • Factor out ((\ln 3)^{x}): (g^\prime(x)=(\ln 3)^{x}(\frac{1}{2\sqrt{x}}+\sqrt{x}\ln(\ln 3))).

Answer:

For (f(x)=(3x^{2}-4x + 5)\cdot2^{x}), (f^\prime(x)=2^{x}(6x - 4 + 3x^{2}\ln 2-4x\ln 2 + 5\ln 2)); for (g(x)=\sqrt{x}\cdot(\ln 3)^{x}), (g^\prime(x)=(\ln 3)^{x}(\frac{1}{2\sqrt{x}}+\sqrt{x}\ln(\ln 3)))