Question

1. (2 points) find the derivative of each function. pick 2.

a. $f(x)=(3x^{2}-4x + 5)cdot2^{x}$
b. $g(x)=sqrt{x}cdot(ln 3)^{x}$
c. $h(x)=\frac{3 + 2sqrt{x}}{7^{x}}$

Explanation:

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 \(2^{x}\): \(f^\prime(x)=2^{x}[(6x - 4)+(3x^{2}-4x + 5)\ln 2]=2^{x}(6x - 4 + 3x^{2}\ln 2-4x\ln 2 + 5\ln 2)\).

1. 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 \(g(x)=\sqrt{x}\cdot(\ln 3)^{x}\), \(g^\prime(x)=(\ln 3)^{x}(\frac{1}{2\sqrt{x}}+\sqrt{x}\ln(\ln 3))\)