The slope of the tangent to the curve `(y-x^5)^2=x(1+x^2)^2`
at the point `(1,3)`
is.
Text Solution
Verified by Experts
`(y-x^(5))^(2)=x(1+x^(2))^(2)` Differentiating both sides w.r.t. x, we get `2(y-x^(5))((dy)/(dx)-5x^(4))=1(1+x^(2))^(2)+(x)(2(1+x^(2))(2x))` On putting x=1 , y =3 in above equation, we get `(dy)/(dx)=8`
Topper's Solved these Questions
DIFFERENTIATION
CENGAGE|Exercise Matrix Match Type|1 Videos
DIFFERENTIAL EQUATIONS
CENGAGE|Exercise Question Bank|25 Videos
DOT PRODUCT
CENGAGE|Exercise DPP 2.1|15 Videos
Similar Questions
Explore conceptually related problems
Find the slope of the tangent to the curve y(x^2+1)=x at the point (1, 1/2)
Find the slope of the tangent to the curve : y=x^(3)-2x+8 at the point (1, 7) .
If m is the slope of the tangent to the curve e^(y)=1+x^(2) , then
Find the angle of inclination of the slope of the tangent to the curve x^(2)+2y=8x-7 at the point x=5.
Slope of the tangent to the curve y=x^(2)+3 at x=2 is
The slope of the tangent line to the curve y=x^3-2x+1 at x=1 is
The slope of the tangent to the curve y=6+x-x^(2) at (2,4) is
