Description
11)Differentiate:f(x)=x^2+2tanx
12)Find an equation for the tangent line to the graph of f(x)=2x^2-2x+3 at the point where x=1
13)Let f(7)=0, f'(7)=14,g(7)=1 and g'(7)=1/7 find h(7) if h(x)=f(x)/g(x)
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Explanation & Answer
11. f1(x)=2x+2sec^2x
12.
take the derivative: 4x -2 (power rule)put 1 in for x, like given in problem.
so... 4(1)-2 = 2
that would be the slope of the line
then to find the point: you have the x-value, 1 (given)
you put 1 in for x in the origianl equation.
2(1)^2 - 2(1) +3 = 2 - 2 +3 = 3
so in point-slope form the answer is:
y - 3 = 2(x - 1)
or
y= 2x+1
13.
h'(x)=[f'(x)*g(x)-g'(x)*f(x)]/[g(x)]^2 {quotient rule}
So h'(7)=[f'(7)*g(7)-g'(7)*f(7)]/[g(7)]^2 {subbing in}
=[(1)*(14)-(0)*(1/7)]/[1]^2
=14
So h'(7)=[f'(7)*g(7)-g'(7)*f(7)]/[g(7)]^2 {subbing in}
=[(1)*(14)-(0)*(1/7)]/[1]^2
=14
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