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anonymous
 4 years ago
the derivate of h(x)=4/7x is 4/7?
anonymous
 4 years ago
the derivate of h(x)=4/7x is 4/7?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jimmy can you help me with this find the slope of the tangent line of f(x)= 2x^2x+3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0best written as (6/7) x though jinnie  so its not confused with 4 divided by 7x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea ok find derivative of f(x) f'(x) = 4x  1  thats it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for example the slope of tangent at x = 1 = 4(1)  1 = 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i must find the derivative first which is 2x1+0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah  but its 4x  1 The derivative at a point on a curve is slope of the tangent at that point

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yea you have to multiply ok i got it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what is the equation of the tangent line at x=1? is it the same as the slope of the tangent line at the point x=1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no use the formula yy1 = m(xx1) to get the equation at x=1 , y = 2(1)^2  1 + 3 = 4 so we need the equation at the point (1,4) with a slope of 3 y  4 = 3 (x  1) is equation of the tangent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can i continue with that or is that where it stops

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can simplify that y  4 = 3x  3 y = 3x + 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the way to find the equation of the tangent at a point on a curve is to 1. differentiate the equation of the curve to find slope of the curve 2 substitute for slope, and x and y values of the point into the general form yy1 = m(xx1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you get the slope of 3 btw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i found the slope in terms of x by differentiating f(x) = 2x^2  x +3 f'(x) = 4x  1  this the slope in terms of x , at point x=x so at x = 1, slope is found by plugging x=1 into 4x1: slope = 4(1)  1 = 3
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