Question

Consider the following. \[ \cos (x)+\sqrt{y}=1 \] (a) Find \( y^{\prime} \) by implicit differentiation. \[ y^{\prime}= \] \( \square \) (b) Solve the equation explicitly for \( y \) and differentiate to get \( y^{\prime} \) in terms of \( x \). \[ y^{\prime}=\square \] (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \( y \) into your solution for part (a). \[ y^{\prime}=\square \]

          Consider the following.
\[
\cos (x)+\sqrt{y}=1
\]
(a) Find \( y^{\prime} \) by implicit differentiation.
\[
y^{\prime}=
\]
\( \square \)
(b) Solve the equation explicitly for \( y \) and differentiate to get \( y^{\prime} \) in terms of \( x \).
\[
y^{\prime}=\square
\]
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \( y \) into your solution for part (a).
\[
y^{\prime}=\square
\]

Consider the following.

cos (x)+√(y)=1

(a) Find y^' by implicit differentiation.

y^'=

□
(b) Solve the equation explicitly for y and differentiate to get y^' in terms of x.

y^'=□

(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a).

y^'=□

Added by Laura M.

Biocalculus Calculus for the Life Sciences

James Stewart 1st Edition

Chapter 3

Instant Answer

Solved by Expert Hoan Nguyen

Step 1

The derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \), and the derivative of \( \sqrt{y} \) with respect to \( x \) is \( \frac{1}{2\sqrt{y}} \cdot y' \) by the chain rule. The derivative of the constant \( 1 \) is \( 0 \). So we have: \[ Show more…

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Consider the following. \[ \cos (x)+\sqrt{y}=1 \] (a) Find \( y^{\prime} \) by implicit differentiation. \[ y^{\prime}= \] \( \square \) (b) Solve the equation explicitly for \( y \) and differentiate to get \( y^{\prime} \) in terms of \( x \). \[ y^{\prime}=\square \] (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \( y \) into your solution for part (a). \[ y^{\prime}=\square \]