Question

Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)= te^11-6t5) g(p) = pIn(2p+3)6) z= (te^6t + e^5t)^77) w= 2y+y^2 / 7+y

Answer 1

Answer:

Since,

$\frac{d}{dx}x^n = nx^{n-1}$

$\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))$

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}$

1) $y = 8x e^x$

Differentiating with respect to x,

$\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)$

2) $y = 5x e^{x^4}$

Differentiating w. r. t x,

$\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)$

3) $y = x^8 + \frac{5}{x^4}$

Differentiating w. r. t. x,

$\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}$

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

$g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)$

6) $z = (te^{6t}+e^{5t})^7$

Differentiating w. r. t. t,

$\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})$

7) $w =\frac{2y + y^2}{7+y}$

Differentiating w. r. t. y,

$\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}$