Implicit differentiation enables us to findfor any relation involving x and y, even if y cannot be solved for explicitly! For example, consider the equation:y + y – 2x = 5a. If possible, solve this equation for y in terms of x. If not possible, explain why not.b. Use implicit differentiation to writedyin terms of x and y.dec. If possible, determine any values of a for which this relation is non-differentiable.d. Does this relation have any horizontal tangents? Usedæto support your claim.e. Calculate the slopes of the tangent lines at the points (2.5, 2) and (-1.5, 1).dyf. What isda2? Simplify your answer so that only y and x are present.

Answered: Implicit differentiation enables us to…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

Implicit differentiation enables us to find
for any relation involving x and y, even if y cannot be solved for explicitly! For example, consider the equation:
y + y – 2x = 5
a. If possible, solve this equation for y in terms of x. If not possible, explain why not.
b. Use implicit differentiation to write
dy
in terms of x and y.
de
c. If possible, determine any values of a for which this relation is non-differentiable.
d. Does this relation have any horizontal tangents? Use
dæ
to support your claim.
e. Calculate the slopes of the tangent lines at the points (2.5, 2) and (-1.5, 1).
dy
f. What is
da2
? Simplify your answer so that only y and x are present.

Expert Solution

Step 1

Given cubic equation is

$y^{3} + y - 2 x = 5$

a. It is not possible to express y in terms of x. because it is a cubic relation in y and relation has a linear term in y too.

we can express x in terms of y as

$x = \frac{1}{2} (y^{3} + y - 5)$

b. Implicit differentiation,

$\begin{array}{rcl} f (x, y) & = & y^{3} + y - 2 x - 5 \\ \frac{d y}{d x} & = & - \frac{\frac{\partial}{\partial x} f (x, y)}{\frac{\partial}{\partial y} f (x, y)} \\ = & - \frac{\frac{\partial}{\partial x} (y^{3} + y - 2 x - 5)}{\frac{\partial}{\partial y} (y^{3} + y - 2 x - 5)} \\ \frac{d y}{d x} & = & \frac{2}{3 y^{2} + 1} \end{array}$

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Single Variable$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions