let\n$f(x)=5x^{5}ln x$\n$f(x)=10x^{4}(ln (x))+5x^{4}$\n$f(e^{3})=10e^{12}(ln (e^{3}))+5e^{12}$
let\n$f(x)=5x^{5}ln x$\n$f(x)=10x^{4}(ln (x))+5x^{4}$\n$f(e^{3})=10e^{12}(ln (e^{3}))+5e^{12}$
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u = 5x^{5}) and (v=\ln x). The derivative of (u = 5x^{5}) with respect to (x) is (u^\prime=5\times5x^{4}=25x^{4}) (using the power rule ((x^{n})^\prime = nx^{n - 1})). The derivative of (v=\ln x) with respect to (x) is (v^\prime=\frac{1}{x}). So, (f^\prime(x)=u^\prime v+uv^\prime=25x^{4}\ln x+5x^{5}\times\frac{1}{x}).
Step2: Simplify the expression
Simplify (5x^{5}\times\frac{1}{x}) to (5x^{4}). So, (f^\prime(x)=25x^{4}\ln x + 5x^{4}=5x^{4}(5\ln x + 1)).
Step3: Evaluate (f^\prime(e^{3}))
Substitute (x = e^{3}) into (f^\prime(x)). We know that (\ln(e^{3})=3) (since (\ln(a^{b})=b\ln a) and (\ln e = 1)). (f^\prime(e^{3})=25(e^{3})^{4}\ln(e^{3})+5(e^{3})^{4}). ((e^{3})^{4}=e^{12}), and (\ln(e^{3}) = 3). (f^\prime(e^{3})=25e^{12}\times3+5e^{12}). (f^\prime(e^{3})=(75 + 5)e^{12}=80e^{12}).
Answer:
(f^\prime(x)=25x^{4}\ln x+5x^{4}), (f^\prime(e^{3}) = 80e^{12})