A. J. Derby Net Worth 2018: What is this NFL football player worth?


A. J. Derby is a professional football player who plays Tight end for the Miami Dolphins as number 85. Derby was born on September 20, 1991, in Iowa City, Iowa. Derby’s height is 6 ft 4 in. This page looks closely at A. J. Derby’s net worth.

A. J. Derby Net Worth 2018: What is this NFL football player worth?Image Credit: Jeffrey Beall

A. J. Derby Career and Earnings/Salary

In high school, A. J. Derby played football for Iowa City High School. Derby played college football on the Arkansas Razorbacks football team. In the 2015 NFL draft, Derby was chosen number 202 in round 6.

In the past, Derby played for the New England Patriots, Denver Broncos, and Miami Dolphins.

Some of Derby’s most prominent statistics include 2 Receiving touchdowns, 404 Receiving yards, and 37 Receptions.

A. J. Derby Net Worth 2018

NFL player salaries range heavily. At the starting level, NFL rookies earn between $400,000 and $600,000 per year. At the highest, stars can earn $50 million plus. The NFL salary cap is around $175 million.

A. J. Derby net worth
A. J. Derby net worth: football salary distribution

So what is NFL football player A. J. Derby’s net worth in 2018? Our estimate for A. J. Derby’s net worth as of 2018 is:

Looking for other NFL football players? Check out these net worth articles: Nick Sebek, Paul Jetton, Ted Ginn Jr., Billy Alford, Selwyn Lymon, La’el Collins, Knute Rockne, John Fuqua, Chuck Bednarik, Perry Griggs, and Kevin Graf.

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