17.Find dydxfor the following functions. Do not simplify your answer.a) y = x4log4(x2 − 5)b) y =ln(sin(x))4xc) x = 7 + exyd) y = sec ( √x)e) y = etan(x3)f) y = (sin(x + 4))x2

a) To find dy/dx for y = x^4 * log4(x^2 - 5), we use the product rule (uv)' = u'v + uv'. Here, u = x^4 and v = log4(x^2 - 5).

u' = 4x^3 v' = 1/(x^2 - 5) * 2x * 1/log(4) (using the chain rule)

So, dy/dx = 4x^3 * log4(x^2 - 5) + x^4 * 1/(x^2 - 5) * 2x * 1/log(4)

b) For y = ln(sin(x))^4/x, we use the quotient rule (u/v)' = (u'v - uv')/v^2. Here, u = ln(sin(x))^4 and v = x.

u' = 4 * ln(sin(x))^3 * 1/sin(x) * cos(x) (using the chain rule) v' = 1

So, dy/dx = (4 * ln(sin(x))^3 * 1/sin(x) * cos(x) * x - ln(sin(x))^4)/x^2

c) For x = 7 + e^(xy), we implicitly differentiate with respect to x.

dx/dx = 1 d/dx [7 + e^(xy)] = 0 + e^(xy) * (y + x * dy/dx) (using the chain rule)

So, 1 = e^(xy) * (y + x * dy/dx). Solve for dy/dx to get the derivative.

d) For y = sec(√x), we use the chain rule (f(g(x)))' = f'(g(x)) * g'(x). Here, f(u) = sec(u) and g(x) = √x.

f'(u) = sec(u) * tan(u) g'(x) = 1/(2√x)

So, dy/dx = sec(√x) * tan(√x) * 1/(2√x)

e) For y = e^(tan(x^3)), we again use the chain rule. Here, f(u) = e^u and g(x) = tan(x^3).

f'(u) = e^u g'(x) = sec^2(x^3) * 3x^2

So, dy/dx = e^(tan(x^3)) * sec^2(x^3) * 3x^2

f) For y = (sin(x + 4))^x^2, we use the chain rule and power rule. Here, f(u) = u^x^2 and g(x) = sin(x + 4).

f'(u) = x^2 * u^(x^2 - 1) g'(x) = cos(x + 4)

So, dy/dx = x^2 * (sin(x + 4))^(x^2 - 1) * cos(x + 4) + (sin(x + 4))^x^2 * 2x * log(sin(x + 4)) (using the product rule on the power function).