the equation of the line tangent to the graph of $f(x)=\\sqrt{10x}$ at $(10,10)$ is $y=\\square$.
the equation of the line tangent to the graph of $f(x)=\\sqrt{10x}$ at $(10,10)$ is $y=\\square$.
Answer
Explanation:
Step1: Find the derivative of (f(x))
Rewrite (f(x)=\sqrt{10x}=(10x)^{\frac{1}{2}}). Using the power - rule ((u^n)^\prime = nu^{n - 1}u^\prime) (where (u = 10x), (n=\frac{1}{2})), (f^\prime(x)=\frac{1}{2}(10x)^{-\frac{1}{2}}\times10=\frac{5}{\sqrt{10x}}).
Step2: Evaluate the derivative at (x = 10)
Substitute (x = 10) into (f^\prime(x)), (f^\prime(10)=\frac{5}{\sqrt{10\times10}}=\frac{5}{10}=\frac{1}{2}). So the slope (m=\frac{1}{2}).
Step3: Use the point - slope form (y - y_1=m(x - x_1))
Given the point ((x_1,y_1)=(10,10)) and (m=\frac{1}{2}), (y - 10=\frac{1}{2}(x - 10)). Expand: (y-10=\frac{1}{2}x - 5). Add (10) to both sides: (y=\frac{1}{2}x+5).
Answer:
(y=\frac{1}{2}x + 5)