Question

Differentiate using substitution method

Answer 1

Answer:

Substitute

$\sqrt{x} = \tan( z ) \\ \sqrt{a} = \tan( \alpha ) \\ \\ y = \tan { }^{ - 1} ( \frac{ \tan(z) + \tan( \alpha ) }{1 - \tan(z) \tan( \alpha ) } ) \\ \\ y = \tan {}^{ - 1} ( \tan(z + \alpha ) ) \\ \\ y = \alpha + z \\ \\ \frac{dy}{dx} = \frac{dz}{dx} = \frac{d}{dx} \tan {}^{ - 1} ( \sqrt{x} ) \\ \\ dy = \frac{1}{2 \sqrt{x} (1 + x)}$

Answer 2

Step-by-step explanation:

\begin{gathered}\sqrt{x} = \tan( z ) \\ \sqrt{a} = \tan( \alpha ) \\ \\ y = \tan { }^{ - 1} ( \frac{ \tan(z) + \tan( \alpha ) }{1 - \tan(z) \tan( \alpha ) } ) \\ \\ y = \tan {}^{ - 1} ( \tan(z + \alpha ) ) \\ \\ y = \alpha + z \\ \\ \frac{dy}{dx} = \frac{dz}{dx} = \frac{d}{dx} \tan {}^{ - 1} ( \sqrt{x} ) \\ \\ dy = \frac{1}{2 \sqrt{x} (1 + x)}\end{gathered}

x

=tan(z)

a

=tan(α)

y=tan

−1

(

1−tan(z)tan(α)

tan(z)+tan(α)

)

y=tan

−1

(tan(z+α))

y=α+z

dx

dy

=

dx

dz

=

dx

d

tan

−1

(

x

)

dy=

2

x

(1+x)

1