Below are few calculus  Questions which  I solved.. 

(1)Find the derivative of the following if:


(i) 2y²=x²siny+xy³
(ii)x²y²+4xy+x-6y=2

(2)differentiate y=cos-¹(4x³-1)

(3)determine the turning or critical point of 

f(t)=(8/3)t³+t²-3t+4

(4)differentiate the function below

SOLUTION 
 
(1)(i) To differentiate this function involves using Implicit function which is defined by the formula dy/dx=-fx/fy, where fx= partial derivative of x and 
fy= partial derivative of y. 
Now equate the whole function to zero. 
 2y²-x²siny-xy³=0
Now we proceed to taking fx
Fx=-2xsiny-y³ and
Fy=4y-x²cosy-3xy²
Using dy/dx=-fx/fy we get
ii) x²y²+4xy+x-6y=2 equate the function to zero
x²y²+4xy+x-6y-2=0
We first take the partial derivative of x i.e
Fx=2xy²+4y+1 and fy=2x²y+4x-6
Now we substitute fx and fy into the Implicit function formula which is 
dy/dx=-fx/fy

(2)  y=x³cos-¹(4x³-1) here we use the product Rule of differentiation which is 
So we let u=x³, and v=cos-¹(4x³-1) so that du/dx=3x²
To differentiate v, we use derivative of  inverse of trigonometric function which is defined below 
From v=cos-¹(4x³-1),  we let p=4x³-1, so that dp/dx=12x², substituting  p and dp/dx into the inverse trigonometric formula yields 

(3)  The given function f(t) can be differentiated, so to obtain the critical or turning points  we simply take the first derivative and equate it to zero. 
f'(t)=8t²+2t-3, to determine the turning or critical point, we let f'(t)=0
8t²+2t-3=0 we now factorize the Quadratic equation, 8t²-4t+6t-3=0
=(8t²-4t)+(6t-3)=0
=>4t(2t-1)+3(2t-1)=0
Hence 4t+3=0
=>t=-¾
And 2t-1= 0 =>
t=½, hence the turning or critical points are t=(-¾,½). 
(4)To differentiate the function 
We use qoutient rule to differentiate the whole function, to use qoutient rule we let the numerator be u, and denominator be v, so that u= (x²+3)²siny........(i) , since u is a composite function, we differentiate u using product rule, so let p=(x²+3)² and q=sin2x
dp/dx=2(x²+3) this is done by using chain Rule. 
dq/dx=2cos2x
du/dx=sin2x(4x(x²+3)+(x²+3)²(2cos2x)
=(4x³+12x)sin2x+2cos2x(x²+3)².......... (ii)   and hence we are done with differentiating the numerator we now move onto differentiating the denominator, let v=(x²+3)1......... (iii) . , dv/dx=2(x+1)........... (iv) 
Now substitute (i), (ii), (iii) and (iv)  into the Qoutient Rule formula below
And this yields