What is the derivative of f(x)=3^(5x^2-4x). I'm shown that the answer is f'(x)=2ln3(5x-2)(3^(5x^2-4x)) but I'm not sure where the 2 in the front is coming from. Instead of 5x-2 shouldnt it also be 10x-4?
+6248 \(f(x) = 3^{5x^2-4x} = e^{\ln(3)(5x^2-4x)}\)
\(\dfrac{d}{dx} g(f(x)) = g^\prime(f(x))f^\prime(x) \\ \text{here we have}\\ g(x) = e^x \\ f(x) = \ln(3)(5x^2-4x)\)
\(\dfrac{d}{dx}e^x = e^x \text{ so }\\ g^\prime(f(x)) = e^{\ln(3)(5x^2-4x)} \\ f^\prime(x) = \ln(3)(10x-4) \text{ so }\\ g^\prime(f(x)) f^\prime(x) = e^{\ln(3)(5x^2-4x)}\cdot \ln(3)(10x-4) = \\ 2\ln(3)(5x-2)e^{\ln(3)(5x^2-4x)} = \\ 2\ln(3)(5x-2)3^{5x^2-4x}\)
ugh.. I posted all that when all I had to say was that
2(5x-2) = 10x - 4
.+6248 \(f(x) = 3^{5x^2-4x} = e^{\ln(3)(5x^2-4x)}\)
\(\dfrac{d}{dx} g(f(x)) = g^\prime(f(x))f^\prime(x) \\ \text{here we have}\\ g(x) = e^x \\ f(x) = \ln(3)(5x^2-4x)\)
\(\dfrac{d}{dx}e^x = e^x \text{ so }\\ g^\prime(f(x)) = e^{\ln(3)(5x^2-4x)} \\ f^\prime(x) = \ln(3)(10x-4) \text{ so }\\ g^\prime(f(x)) f^\prime(x) = e^{\ln(3)(5x^2-4x)}\cdot \ln(3)(10x-4) = \\ 2\ln(3)(5x-2)e^{\ln(3)(5x^2-4x)} = \\ 2\ln(3)(5x-2)3^{5x^2-4x}\)
ugh.. I posted all that when all I had to say was that
2(5x-2) = 10x - 4
