Find the exact value of the gradient of the curve y = e^(2- x)ln(3x- 2). at the point on the curve where x = 2.
To solve this problem, we must first differentiate:
Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2).
Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)
By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).
When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.
JC
Related Maths A Level answers
All answers ▸The curve C has the equation y = 2e^x -6lnx and passes through the point P with x - coordinate 1. a) Find the equation to the tangent to C at P
Find the coordinates of the stationary point on the curve y=2x^2+3x+4=0
y = 3x^2 + 2x^(1/2) - 12 Find dy/dx
Express 9^(3x+1) in the form 3^y, giving "y" in the form "ax+b" where "a" and "b" are constants.