Question

differentiate the following expressions

1fbf0a05f46c8850ce2e9d1d9f0bfc97.txt

Answer 1

Your answer is here :-[

Step-by-step explanation:

Find the derivative of the following via implicit differentiation:

d/dx(y) = d/dx(sqrt(1 + x/sqrt(1 - x)))

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):

d/dx(x) y'(x) = d/dx(sqrt(1 + x/sqrt(1 - x)))

The derivative of x is 1:

1 y'(x) = d/dx(sqrt(1 + x/sqrt(1 - x)))

Using the chain rule, d/dx(sqrt(x/sqrt(1 - x) + 1)) = ( dsqrt(u))/( du) ( du)/( dx), where u = x/sqrt(1 - x) + 1 and d/( du)(sqrt(u)) = 1/(2 sqrt(u)):

y'(x) = (d/dx(1 + (1 - x)^(-1/2) x))/(2 sqrt(1/sqrt(1 - x) x + 1))

Differentiate the sum term by term:

y'(x) = d/dx(1) + d/dx((1 - x)^(-1/2) x) 1/(2 sqrt(1 + x/sqrt(1 - x)))

The derivative of 1 is zero:

y'(x) = (d/dx(x/sqrt(1 - x)) + 0)/(2 sqrt(1 + x/sqrt(1 - x)))

Simplify the expression:

y'(x) = (d/dx((1 - x)^(-1/2) x))/(2 sqrt(1 + x/sqrt(1 - x)))

Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = 1/sqrt(1 - x) and v = x:

y'(x) = x d/dx((1 - x)^(-1/2)) + (d/dx(x))/(sqrt(1 - x)) 1/(2 sqrt(1 + x/sqrt(1 - x)))

Using the chain rule, d/dx(1/sqrt(1 - x)) = d/( du)1/sqrt(u) ( du)/( dx), where u = 1 - x and d/( du)(1/sqrt(u)) = -1/(2 u^(3/2)):

y'(x) = ((d/dx(x))/sqrt(1 - x) + -(d/dx(1 - x))/(2 (1 - x)^(3/2)) x)/(2 sqrt(1 + x/sqrt(1 - x)))

Differentiate the sum term by term and factor out constants:

y'(x) = ((d/dx(x))/sqrt(1 - x) - d/dx(1) - d/dx(x) x/(2 (1 - x)^(3/2)))/(2 sqrt(1 + x/sqrt(1 - x)))

The derivative of 1 is zero:

y'(x) = ((d/dx(x))/sqrt(1 - x) - (x (-(d/dx(x)) + 0))/(2 (1 - x)^(3/2)))/(2 sqrt(1 + x/sqrt(1 - x)))

Simplify the expression:

y'(x) = ((d/dx(x))/sqrt(1 - x) + (x (d/dx(x)))/(2 (1 - x)^(3/2)))/(2 sqrt(1 + x/sqrt(1 - x)))

The derivative of x is 1:

y'(x) = ((d/dx(x))/sqrt(1 - x) + 1 x/(2 (1 - x)^(3/2)))/(2 sqrt(1 + x/sqrt(1 - x)))

The derivative of x is 1:

y'(x) = (x/(2 (1 - x)^(3/2)) + 1 1/sqrt(1 - x))/(2 sqrt(1 + x/sqrt(1 - x)))

Factor the numerator and denominator of the right hand side:

y'(x) = (-2 + x)/((2 sqrt(1 - x) (-1 + x)) (2 sqrt(1 + x/sqrt(1 - x))))

Cancel common terms in the numerator and denominator:

Answer: |

| y'(x) = (-2 + x)/(4 sqrt(1 - x) (-1 + x) sqrt(1 + x/sqrt(1 - x)))

Answer 2

Answer:

Step-by-step explanation:

(1) $\frac{d}{dx}(1/x)$  = ln(x)

(2) $\frac{d}{dx}(x(x+1)$

⇒$\frac{d}{dx}(x^2 + x)$

⇒$2x + 1$