something wrong with my solution, need help!
- Thread starter kahheng
- Start date
Unfortunately, your very first line is wrong. The logarithm of a sum doesn't simplify to the sum of the logarithms. In fact, the sum of the logarithms is the log of the product. You can't take advantage of logs when you have a sum of terms.Differentiate
\(\displaystyle y = 2^x + 2(3^x) \)
\(\displaystyle \dfrac{dy}{dx} = \dfrac{d}{dx}(2^x) + \dfrac{d}{dx}[2(3^x)] \)
Give that a try, and let us know if you need more help.
Your problem occurs in the third line.Differentiate
\(\displaystyle y = f(x) = 2^x + 2(3^x) \implies ln(y) = ln\{2^x + 2(3^x)\} \ne ln(2x) + ln(2 * 3^x).\)
You have made this harder than it needs to be
\(\displaystyle u = 2^x\ and\ v = 3^x \implies y = u + 2v \implies \dfrac{dy}{dx} = \dfrac{du}{dx} + 2 * \dfrac{dv}{dx}.\)
Now \(\displaystyle w = a^x \implies ln(w) = ln(a^x) = xln(a) \implies \dfrac{1}{w} * \dfrac{dw}{dx} = ln(a) * 1\implies \dfrac{1}{a^x} * \dfrac{dw}{dx} = ln(a) \implies \dfrac{dw}{dx} = ln(a) * a^x.\)
\(\displaystyle So\ \dfrac{du}{dx} = ln(2) * 2^x\ and\ \dfrac{dv}{dx} = ln(3) * 3^x \implies\)
\(\displaystyle \dfrac{dy}{dy} = ln(2) * 2^x\ + 2 * ln(3) * 3^x.\)
You messed up in the algebra, not the calculus. Happens to me too frequently.