Type your question here 6. 6 points Previous Answers My Notes SE 22 If f(x) Cf(x
ID: 3342201 • Letter: T
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6. 6 points Previous Answers My Notes SE 22 If f(x) Cf(x)] 18 and f(1) 32, find f (1 f '(1 Enhanced Feedback Please try again using implicit differentiation. Differentiate both sides with respect to x. Use the Chain Rule or Product Rule when necessary (terms involving f(x) will always need the Chain Rule). Evaluate the terms for the given calcPad g the g Th for f quati Need Help? Read It Watch It Chat About It DNE Functions 6 points My Notes I ET2 2.6.019 Use implicit differentiation to find an equation of the tangent line to the curve at the given point 7x2 xy 7 15, (1, 1) (ellipse) Set Greek Hen Need Help? Read It Watch It Chat About It 8. 6 points l Previous Answers My Notes I ET2 20 Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2 2xy y x 2, 1, 2 hyperb 2 Enhanced Feedback Please try again. In order to find the equation of the tangent line, you first have to find the slope of the tangent line at the given point. To do so, you have to use implicit differentiation. Differentiate both sides with respect to x. Use the Chain Rule or Product Rule when necessary (terms involving y will always need the Chain Rule). Evaluate the terms for the given point (a, b). Let x 3a and y 3 b. Then solve the equation for y. This will give you the slope of the tangent line. Use the point-slope form to find the equation of the tangent line Need Help? Read It Chat About It 9. 6 points I Previous Answers My Notes I ET2 2.6.021 Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 4x2 2y2 (0, 0.5 (cardioid)Explanation / Answer
f(x)+x^2[f(x)]^4=18
differentiate on bothsides
f'(x)+2x(f(x))^4+3f(x)^3f'(x)x^2=0
sub stitute x=1 and f(1)=2
f'(1)+2[f(1)^4]+3f(1)^3f'(1)=0
f'(1)+2*2^4+3*2*f'(1)=0
f'(1)=-32/7
B.7x^2+7y^2+xy=15
differentiate on bothsides
14x+14yy'+xy'+y=15
substitute x=1 y =1
14+14y'+y'+1=15
y'=0
slope of tangent is zero
y=k
is tangent is eq
k=1 as it is passes through 1,1 point
y=k tangent
C.
x^2+2xy-y^2+x=2
differentiate on bothsides
2x+2xy'+2y-2yy'+1=0
at point(1,2)
2+2y'+4-4y'+1=0
-2y'=-7
y'=7/2 =slope of tangent =m
y-y1/x-x1=m
y-1/x-2=7/2
2y-2=7x-14
7x-2y-12=0
tangwnt is y=7/2x-3/2
2y=7x-3
D.
x^2+y^2=(4x^2+2y^2-x)^2
differentiate on bothsides
2x+2yy'=2(x^2+2y^2-x)(8x+4yy'-1)
substiyue x=0.y=0.5
1y'=2(2(0.25)(2y'-1)
0.5y'=2y'-1
y'=2/3=m=slope
y-y1/x-x1=m
y1=0.5
x1=0
m=2/3
y-0.5/x=2/3
3y-1.5=2x
2x-3y+1.5=0
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